A phase transition in the evolution of bootstrap percolation
processes on preferential attachment graphs
Mohammed Amin Abdullah∗
Nikolaos Fountoulakis†
February 6, 2017
Abstract
The theme of this paper is the analysis of bootstrap percolation processes on random
graphs generated by preferential attachment. This is a class of infection processes where
vertices have two states: they are either infected or susceptible. At each round every
susceptible vertex which has at least r ≥ 2 infected neighbours becomes infected and
remains so forever. Assume that initially a(t) vertices are randomly infected, where t
is the total number of vertices of the graph. Suppose also that r < m, where 2m is
the average degree. We determine a critical function ac (t) such that when a(t) ac (t),
complete infection occurs with high probability as t → ∞, but when a(t) ac (t), then
with high probability the process evolves only for a bounded number of rounds and the
final set of infected vertices is asymptotically equal to a(t).
1
Introduction
The dissemination of contagion within a network is a fundamental problem that arises in a
wide spectrum of social and economic sciences. Among the mechanisms which underlie this
phenomenon is a class of dissemination processes where local decisions (or microbehaviours)
aggregate into a large outbreak or pandemic. Quite frequently, these phenomena begin on a
rather small scale and may end up contaminating a large part of the network. What are the
particular characteristics of a network that enable or inhibit such an outbreak?
A general class of models that incorporates this kind of behaviour is what is called the
general threshold model [31]. Here it is assumed that each vertex has one of two states: it is
either infected or susceptible. Furthermore, each vertex of the underlying graph is equipped
with a threshold function which depends on the states of its neighbours. This function
expresses the probability that this vertex remains in a particular state given the states of its
neighbours. A central problem in viral marketing is given a network, a parameter k ≥ 1 and
such a set of functions, find a set of vertices S of size k which maximizes the expected number
of infected vertices at the end of the process. In [29], Kempe, Kleinberg and Tardos proved
that finding such an optimal set is NP-hard. Moreover, they showed that it is NP-hard to
approximate the size of the maximum expected outreach even within a polynomial factor.
See also [30] for similar results.
∗
Mathematical and Algorithmic Sciences Lab, Huawei Technologies Ltd. [email protected]
This work was mostly conducted whilst M. A. Abdullah was at the University of Birmingham.
†
School of Mathematics, University of Birmingham, Edgbaston, B15 2TT, U.K. [email protected]
Research supported by the EPSRC Grant No. EP/K019749/1.
1
In this paper, we study an instance of this class of models known as bootstrap percolation
processes. This is a threshold model that was introduced in the context of mathematical
physics by Chalupa, Leath and Reich [17] in 1979 for magnetic disordered systems.
A bootstrap percolation process with activation threshold an integer r ≥ 2 on a (multi)graph
G = G(V, E) is a deterministic process. Initially, there is a subset I0 = I(0) ⊆ V of
infected vertices, whereas every other vertex is susceptible. This set can be selected either
deterministically or randomly. The process evolves in rounds, where in each round, if a
susceptible vertex has at least r edges connected to infected neighbours, then it also becomes
infected and remains so forever. This is repeated until no more vertices become infected. We
denote the final infected set by If . We denote the set of susceptible (infected) vertices at
round τ in the process by S(τ ) (respectively, I(τ )). Thus, S(τ ), I(τ ) form a partition of the
vertex set V , and If = I(∞). Of course, the above definition makes also perfect sense when
r = 1 – in this case If coincides with the set of vertices of the union of those components of
G which contain vertices in I0 .
Such processes (as well as several variations of them) have been used as models to describe
several complex phenomena in diverse areas, from jamming transitions [36] and magnetic
systems [33] to neuronal activity [5, 22]. Bootstrap percolation processes also have connections
with the dynamics of the Ising model at zero temperature [23], [32]. These processes have
also been studied on a variety of graphs, such as trees [10, 24], grids [16, 26, 8], lattices
on the hyperbolic plane [34], hypercubes [7], as well as on several distributions of random
graphs [4, 11, 28]. Ebrahimi et al. [21] showed that given a graph G, a subset of vertices V 0
and a positive parameter s, the question whether the final set of the bootstrap percolation
process starting at V 0 has size at least s is P-complete. A short survey regarding applications
of bootstrap percolation processes can be found in [3]. The theme of this paper is the study
of bootstrap percolation processes on a preferential attachment random graph on t vertices,
which we denote by PAt (m, δ).
2
Preferential attachment graphs
The preferential attachment models have their origins in the work of Yule [37], where a
growing model is proposed in the context of the evolution of species. A similar model was
proposed by Simon [35] in the statistics of language. The principle of these models was used
by Barabási and Albert [12] to describe a random graph model where vertices arrive one by
one and each of them throws a number of half-edges to the existing graph. Each half-edge is
connected to a vertex with probability that is proportional to the degree of that vertex. This
model was defined rigorously by Bollobás, Riordan, Spencer and Tusnády [14] (see also [13]).
We will describe the most general form of the model which is essentially due to Dorogovtsev
et al. [19] and Drinea et al. [20]. Our description and notation below follow that from the
book of van der Hofstad [25].
The random graph PAt (m, δ) is parameterised by two constants: m ∈ N, and δ ∈ R,
δ > −m. It gives rise to a random graph sequence (i.e., a sequence in which each member
is a random graph), denoted by (PAt (m, δ))∞
t=1 . The tth term of the sequence, PAt (m, δ) is
a graph with t vertices and mt edges. Further, PAt (m, δ) is a subgraph of PAt+1 (m, δ). We
define PAt (1, δ) first, then use it to define the general model PAt (m, δ) (the Barabási-Albert
model corresponds to the case δ = 0).
The random graph PA1 (1, δ) consists of a single vertex with one self-loop. We denote the
2
(1)
(1)
(1)
(1)
vertices of PAt (1, δ) by {v1 , v2 , . . . , vt }. We denote the degree of vertex vi in PAt (1, δ)
by Di (t). Then, conditionally on PAt (1, δ), the growth rule to obtain PAt+1 (1, δ) is as follows:
(1)
(1)
We add a single vertex vt+1 having a single edge. The other end of the edge connects to vt+1
1+δ
t(2+δ)+(1+δ) , and connects to
(1)
(1)
write vt+1 → vi . Any self-loop
itself with probability
Di (t)+δ
t(2+δ)+(1+δ)
(1)
a vertex vi
∈ PAt (1, δ) with probability
– we
at a vertex adds 2 to the degree of that
vertex. For any t ∈ N, let [t] = {1, . . . , t}. Thus,
(
1+δ
for i = t + 1,
(1)
(1)
t(2+δ)+(1+δ)
P vt+1 → vi | PAt (1, δ) =
Di (t)+δ
for i ∈ [t]
t(2+δ)+(1+δ)
The model PAt (m, δ), m > 1, with vertices {1, . . . , t} is derived from PAmt (1, δ/m) with
(1) (1)
(1)
vertices {v1 , v2 , . . . , vmt } as follows: For each i = 1, 2, . . . , t, we contract the vertices
(1)
(1)
(1)
{v(i−1)+1 , v(i−1)+2 , . . . , v(i−1)+m } into one super-vertex, and identify this super-vertex as i in
PAt (m, δ). When a contraction takes place, all loops and multiple edges are retained. Edges
shared between a set of contracted vertices become loops in the contracted super-vertex.
Thus, PAt (m, δ) is a graph on [t].
The above process gives a graph whose degree distribution follows a power law with
exponent 3 + δ/m. This was suggested by the analyses in [19] and [20]. It was proved
rigorously for integral δ by Buckley and Osthus [15]. For a full proof for real δ see [25]. In
particular, when −m < δ < 0, the exponent is between 2 and 3. Experimental evidence
has shown that this is the case for several networks that emerge in applications (cf. [2]).
Furthermore, when m ≥ 2, then PAt (m, δ) is whp connected, but when m = 1 this is not the
case, giving rise to a logarithmic number of components (see [25]).
We describe an alternative, though equivalent, direct construction of (PAt (m, δ))∞
t=1 . Let
PA1 (m, δ) be a single vertex with label 1, having m loops. Given PAt−1 (m, δ), t ≥ 2, the
construction of PAt (m, δ) is as follows: To add vertex t to the graph, we split time step t into
m sub-steps, adding one edge sequentially in each sub-step. For j = 1, 2, . . . , m, denote the
graph after the jth sub-step of time t by PAt,j (m, δ). Hence PAt (m, δ) ≡ PAt,m (m, δ). For
notational convenience, let PAt,0 (m, δ) = PAt−1 (m, δ).
Denote the jth edge added by ej . One end of ej will be attached to vertex t and the other
end will be attached randomly to another vertex (which may be t). Let g(t, j) be the random
variable representing this vertex. For j = 1, 2, . . . , m, let Di (t, j) be the degree of vertex i
in PAt,j (m, δ). That is, for j = 1, 2, . . . , m, Di (t, j) is the degree of vertex i after both ends
of ej have been attached. Furthermore, for notational convenience, let Dt (t, 0) = 0 and for
i ∈ [t − 1], let Di (t, 0) = Di (t − 1).
Now, for j = 1, 2, . . . , m, conditionally on PAt,j−1 (m, δ), PAt,j (m, δ) is generated according
to the following probability rules:
(
Dt (t,j−1)+1+jδ/m
for i = t,
(2m+δ)(t−1)+2j−1+jδ/m
P (g(t, j) = i | PAt,j−1 (m, δ)) =
.
Di (t,j−1)+δ
for
i
∈
[t
−
1]
(2m+δ)(t−1)+2j−1+jδ/m
It is not difficult to see that these two constructions give rise to the same probability
distribution over realisations of (PAt (m, δ))∞
t=1 . It will be sometimes convenient to refer to
one form over the other.
3
2.1
Results
Here as well as in the rest of the paper the term with high probability (whp) means with
probability 1 − o(1) in the space of PAt (m, δ), as t → ∞. We will be using the same term for
events over the product space between PAt (m, δ) and the choice of I0 on [t].
p
Let Xt be a random variable on the above product space. If a ∈ R, we write that Xt → a
(Xt converges to a in probability) if for any ε > 0 we have P (|Xt − a| > ε) → 0 as t → ∞.
The selection of I0 is random and each vertex is infected initially with probability p =
p(t) = a(t)/t, independently of any other vertex. Hence, if a(t) → ∞ as t → ∞, the size of
I0 /a(t) converges in probability to 1.
Recently, Ebrahimi et al. [21] investigated a threshold phenomenon that occurs in the
evolution of the process on a variant of the preferential attachment model, that is very similar
(though not identical) to PAt (m, δ). In our context, their results can be stated as follows. Let
m
γ = 2m+δ
. If a(t) t1−γ log t, then whp If = [t], that is, we have complete infection. They
also identified a subcritical range for a(t). Assume first that rγ ≥ 1. If a(t) t1−γ , then
whp If = I0 , that is no evolution occurs. Now, if rγ < 1, then the same holds but provided
that a(t) t1−1/r . Since γ < 1/r, that is, 1 − γ > 1 − 1/r, it follows that this function is
asymptotically smaller than the t1−γ . Similar results were obtained by the two authors in [1]
for PAt (m, δ).
In this paper, we complete the landscape and show that a critical phenomenon occurs
“around” the function t1−γ =: ac (t) = ac . Our results show that when a(t) ac (t), there is
complete infection whp, but if a(t) ac (t) then either there is no evolution of the process
or it halts in a bounded number of rounds. (In fact, for r = 2 we show a slightly weaker
result that requires a(t) ≤ ac (t)/ log t.) Theorems 1 and 2(i) recover the results of Ebrahimi
et al. [21], but Theorem 2(ii) covers also the case where rγ ≤ 1, closing the gap between the
two threshold functions that were identified by Ebrahimi et al. We should also point out that
Ebrahimi et al. [21] achieve strong probability bounds, that is, the related events occur with
probability that tends to 1 polynomially fast. In this case, the process accumulates only a
small number of infections beyond those incurred initially, so that If is almost equal to I0 .
Inside the critical window, that is, if a(t) = Θ(ac (t)), then with probability asymptotically
bounded away from zero there is complete infection, and with probability bounded away from
zero we have similar behaviour as for the a(t) ac (t) case.
1
Let ω = ω(t) → ∞ as t → ∞ arbitrarily slowly. With γ = 2+δ/m
and ac (t) = t1−γ , the
above can be formalized as follows.
Theorem 1 (Supercritical case). If r < m and a(t) = ωac (t) then all vertices in PAt (m, δ)
get infected whp.
Theorem 2 (Subcritical case). If r ≤ m then the following hold:
(i) If a(t) = ac (t)/ω and rγ > 1, then whp, If = I0 .
p
(ii) If a(t) = ac (t)/ω and r ≥ 3 then |If |/|I0 | → 1 and whp the process stops in at most b γ1 c
rounds.
p
(iii) If a(t) = ac (t)/ log t and r = 2, then |If |/|I0 | → 1 and whp the process stops in at most
b γ1 c + 1 rounds.
4
It should be noted that when δ < 0, rγ > 1 is always satisfied, since we insist that r ≥ 2.
Theorem 3 (Critical case). Let r ≥ 3 and a(t) = λac (t) where λ is a constant. Then there
exist constants 0 < p1 , p2 depending on λ such that the following hold:
(i) if r ≤ m, then the following holds with probability at least p1 : vertices are infected for at
most b γ1 c rounds, and |If |/|I0 | < 1 + ε, for any ε > 0 and any t large enough.
(ii) if r < m, then with probability at least p2 , there is a complete infection.
The function ac (t) was also identified by the second author and Amini [6] in the case of
inhomogeneous random graphs of rank 1. However, results of Amini [4] imply that if the
kernel of such a random graph gives rise to a power law degree distribution with exponent
larger than 3 (corresponds to δ > 0), then whp, a sublinear initial infection only results
in a sublinear outbreak. As our results and the results in [21] show this is not the case
in the preferential attachment model. In other words, a sublinear initial infection leads to
an outbreak where every vertex becomes infected, provided that the amount of the initial
infection is not too small. Theorems 1 and 2 identify this critical amount.
Lack of outbreak is also the case in random regular graphs of constant degree [11] as well
as in binomial random graphs with constant expected degree [28]. In the latter case, the
p
authors show that if a(t) = o(t), then |If |/|I0 | → 1. This behaviour is radically different
from that in the preferential attachment model, where Theorem 1 implies that a sublinear
initial infection may lead to pandemics.
2.1.1
The cases r = m and r > m
It can be shown that there are a logarithmic number of self-loops in PAt (m, δ). For r = m,
these loops make analysis of the outcome difficult. This is a rather specific artifact of the
model and, is not shared with slight variations of the model, e.g., one in which self-loops are
not allowed.
For r > m the following “folklore” argument shows that if the number of initially infected
vertices is sublinear, then the final number will be sublinear as well: Let G be the subgraph
induced by all the vertices in If . The number of edges in G is at least (|If | − |I0 |)r but at the
same time, the total number of edges in G can be at most m|If |. Therefore (|If | − |I0 |)r ≤
r
m|If | implying |If | ≤ r−m
|I0 |.
2.2
Further notation and terminology
1
Throughout this paper we let γ = γ(m, δ) = 2+δ/m
, hence 1 − γ = 1+δ/m
2+δ/m . Observe the
condition δ > −m (which must be imposed), implies 0 < γ < 1. Furthermore, δ < 0 if and
only if 21 < γ < 1.
For integers i, j with i ≤ j, we shall sometimes write [i, j] to denote the set {i, i+1, . . . , j}.
We also use Si (t) to denote the sum of degrees for vertices in the interval [1, i], i.e., Si (t) =
P
i
j=1 Dj (t).
We will sometimes say a vertex j throws an edge e to vertex i if, in the construction of
PAj (m, δ), vertex j connected edge e to vertex i. We will also say i receives the edge e.
5
Furthermore, for two non-negative functions f (t), g(t) on N we write f (t) . g(t) to denote
that f (t) = O(g(t)). If, in addition, g(t) = O(f (t)), then we write f (t) g(t). In this paper,
the underlying asymptotic variable will always be t, the number of vertices in PAt (m, δ).
(m,δ)
We use the notation f (c) . g(c) to mean that there is a constant C(m, δ) such that
f (c) ≤ C(m, δ)g(c), and C(m, δ) depends only on m, δ.
We will begin with some general results in the next section on the concentration of the
degrees, which will be used mainly in the Proof of Theorem 1.
3
Vertex degrees: expectation and concentration
As we mentioned above, the degrees in PAt (m, δ) roughly follow a power-law degree distribution with exponent 3 + δ/m, that is, the empirical probability mass function on the degrees
1
scales like x3+δ/m
. In fact, many networks that emerge in applications have a degree distribution that follows a power law with exponent between 2 and 3 (cf. [2] for example), which
corresponds to δ/m ∈ (−1, 0). The Barabasi-Albert model gives power-law with exponent
3 (δ = 0). Observe that the variance on the degrees is finite if and only if the exponent is
greater than 3 (corresponding to δ > 0).
Consider two vertices i and j; their total weight is Di (t) + Dj (t) + 2δ, meaning probability
of an edge being thrown to them is proportional to this value. Now a vertex with degree
Di (t) + Dj (t) would have weight Di (t) + Dj (t) + δ. Thus, we cannot treat two separate
vertices i and j as a single one of the combined degree, except when δ = 0. In the special
case that δ = 0, the weight of a vertex is proportional to its degree, and the weight of a
set of vertices is proportional to the sum of their degrees. When δ = 0, we can treat a set
of vertices as a bucket of half-edges, or stubs, conceptually distributing the stubs across the
vertices however we like. However, when δ 6= 0, the weighting is non-linear. Conceptually
grouping stubs together means that one has to sum their weights not their degrees.
In summary, the probability of a vertex receiving the next edge thrown is proportional to
its weight. The same holds for a set of vertices; the probability a set of vertices receiving an
edge is proportional to the total weight of the set. When, and only when, δ = 0, then the
weight of a vertex is its degree, and the weight of a set is the total degree of the vertices in
the set.
A number of results on the degree sequence are collected in van der Hofstad [25] which
γ
shows, amongst other things, that E[Di (t)] = (1 + o(1))a ti
where a is a constant that
depends only on m and δ.
3.1
Sum of degrees
Recall that Si (t) denotes the sum of degrees for vertices in the interval [1, i]. We state the
following without proof. It is a simple consequence of results in, e.g., [25].
Proposition 4. There exist constants C` , Cu > 0 that depend only on m and δ such that for
each vertex i ∈ [t],
C` tγ i1−γ ≤ E[Si (t)] ≤ Cu tγ i1−γ .
We next derive a concentration results for the sum of degrees. Lemma 5 is an elaboration
of Lemma 2 in [18]. Its proof can be found in the appendix.
6
Lemma 5. Suppose δ ≥ 0. There exists constants K0 , h0 > 0 that depend only on m and δ,
such that the following holds for all i ∈ [t], K > K0 and h < h0 ,
1
P Si (t) < E[Si (t)] ≤ e−hi
K
Lemma 6. Let ε > 0 be a constant. If δ < 0, then there exists a positive constant c = c(m, δ, ε)
that depends only on m, δ and ε, such that with probability at least 1 − e−ci ,
Si (t) ≥ (1 − ε)E[Si (t)]
(1)
for all i ∈ [t].
Proof. We will use a Doob martingale in conjunction with the Azuma-Hoeffding inequality.
(m,δ)
(m,δ)
Define Mn
(i, t) = E[Si (t) | PAn (m, δ)]. Observe, for n = 1, 2, . . . , i, Mn
(i, t) = E[Si (t)].
(m,δ)
(m,δ)
Now we want to bound |Mn+1 (i, t)−Mn
(i, t)| for n ≥ i. Observe that Si (n) is measurable
with respect to PAn (m, δ), and E[Si (t) | Si (n), PAn (m, δ)] = E[Si (t) | Si (n)], i.e., that the
expectation of Si (t) is independent of PAn (m, δ) given Si (n). Hence, we will instead write
(m,δ)
Mn
(i, t) = E[Si (t) | Si (n)]. We have, for t > n,
E[Si (t) + δi | Si (n)] = E[E[Si (t) + δi | Si (t − 1), Si (n)] | Si (n)]
= E[E[Si (t) + δi | Si (t − 1)] | Si (n)].
We will analyse the m = 1 case first. Considering the inner conditional expectation,
E[Si (t) + δi | Si (t − 1)] = Si (t − 1) + δi +
=
Si (t − 1) + δi
(2 + δ)(t − 1) + 1 + δ
(2 + δ)t
(Si (t − 1) + δi) .
(2 + δ)(t − 1) + 1 + δ
Therefore,
E[Si (t) + δi | Si (n)] =
t
t−1+
1+δ
2+δ
E[Si (t − 1) + δi | Si (n)]
t−1
Y
k+1
= (Si (n) + δi)
k + 1+δ
2+δ
k=n
1+δ
= (Si (n) + δi)
Γ(t + 1) Γ(n + 2+δ )
.
Γ(n + 1)
Γ(t + 1+δ
2+δ )
Consequently,
(1,δ)
(1,δ)
Mn+1 (i, t) − Mn (i, t) = |E[Si (t) | Si (n + 1)] − E[Si (t) | Si (n)]|
1+δ Γ(n + 1 + 1+δ
)
Γ(n
+
)
Γ(t + 1) 2+δ
2+δ =
− (Si (n) + δi)
(Si (n + 1) + δi)
Γ(n
+
2)
Γ(n
+
1) Γ(t + 1+δ
)
2+δ
1+δ
1+δ
n + 2+δ
Γ(t + 1) Γ(n + 2+δ ) =
(S
(n
+
1)
+
δi)
−
(S
(n)
+
δi)
.
i
i
Γ(n
+
1)
n
+
1
Γ(t + 1+δ
)
2+δ
7
n+ 1+δ
n
2+δ
We have n+1
< n+1
< 1 and Si (n) ≤ Si (n + 1) ≤ Si (n) + 1, so
n + 1+δ
n + 1+δ
n + 1+δ
2+δ
2+δ
2+δ
− (Si (n) + δi) ≤ (Si (n) + δi) − 1 +
(Si (n + 1) + δi)
n+1
n+1
n+1
<
Si (n) + δi
+ 1.
(2 + δ)(n + 1)
Since Si (n) ≤ 2i + n − i = n + i and i ≤ n, the right-hand side is at most 2:
Si (n) + δi
n + i(1 + δ)
n + n(1 + δ)
≤
≤
< 1.
(2 + δ)(n + 1)
(2 + δ)(n + 1)
(2 + δ)(n + 1)
Thus,
1+δ
Γ(t + 1) Γ(n + 2+δ )
(1,δ)
.
Mn+1 (i, t) − Mn(1,δ) (i, t) < 2
Γ(n + 1)
Γ(t + 1+δ
2+δ )
Recall that when m ≥ 1 we define PAt (m, δ) in terms of PAmt (1, δ/m), and Sa (b) in the
1
former corresponds to Sma (mb) in the latter. Therefore, with γ = γ(m, δ) = 2+δ/m
,
(m,δ)
(1,δ/m)
(1,δ/m)
(mi, mt)
Mn+1 (i, t) − Mn(m,δ) (i, t) = Mm(n+1) (mi, mt) − Mmn
m X
(1,δ/m)
(1,δ/m)
= Mm(n+1)−k+1 (mi, mt) − Mm(n+1)−k (mi, mt) ≤
k=1
m X
k=1
(1,δ/m)
(1,δ/m)
Mm(n+1)−k+1 (mi, mt) − Mm(n+1)−k (mi, mt)
m
≤ 2
Γ(mt + 1) X Γ(m(n + 1) − k + 1 − γ)
.
Γ(mt + 1 − γ)
Γ(m(n + 1) − k + 1)
k=1
We have
Γ(mn + k − γ)
Γ(mn + k)
=
≤
so
mn + k − 1 − γ mn + k − 2 − γ
mn + 1 − γ Γ(mn + 1 − γ)
...
mn + k − 1
mn + k − 2
mn + 1
Γ(mn + 1)
Γ(mn + 1 − γ)
,
Γ(mn + 1)
m
X
Γ(m(n + 1) − k + 1 − γ)
k=1
Γ(m(n + 1) − k + 1)
=
m
X
Γ(mn + k − γ)
Γ(mn + k)
k=1
≤m
Γ(mn + 1 − γ)
.
Γ(mn + 1)
Therefore,
(m,δ)
Mn+1 (i, t) − Mn(m,δ) (i, t) ≤ 2m
Γ(mt + 1) Γ(mn + 1 − γ)
.
Γ(mt + 1 − γ) Γ(mn + 1)
Re-writing the above, we get
Γ(mt + 1 − γ + γ) Γ(mn + 1 − γ)
(m,δ)
(m,δ)
M
(i,
t)
−
M
(i,
t)
n+1
≤ 2m
n
Γ(mt + 1 − γ) Γ(mn + 1 − γ + γ)
γ
t
,
≤ Cm,δ
n
8
where Cm,δ is a universal constant that depends only on m and δ.
Now, applying the Hoeffding-Azuma inequality,


P (Si (t) − E[Si (t)] < −d) ≤ exp 
Since δ < 0, we have
Pt
j=i+1
2γ
t
j
−d2
2
Cm,δ
Pt
j=i+1


2γ  .
t
j
≤ K1 t2γ i1−2γ for some constant K1 .
Hence letting d = εE[Si (t)] ≥ εC` tγ i1−γ for some constant ε > 0,
!
−ε2 C`2 t2γ i2(1−γ)
P (Si (t) − E[Si (t)] < −d) ≤ exp
≤ e−ci
2 K t2γ i1−2γ
Cm,δ
1
for some constant c = c(m, δ, ε) > 0 that depends only on m, δ and ε.
4
Supercritical Case: Proof of Theorem 1
The proof of this theorem relies on the fact that with high probability all of the early vertices
of PAt (m, δ) become infected during the first round. Subsequently, the connectivity of the
random graph is enough to spread the infection to the remaining vertices. The infection of
the early vertices requires sufficiently high lower bounds on their degrees. We show these
using the concentration results of the previous section together with a coupling with a Pólya
urn process.
4.1
Pólya Urns
Consider the following Pólya urn process with red and black balls. Let i ≥ 2 be an integer
and let the weighting functions for the red and black balls be WR (k) = k + δ and WB (k) =
k + (i − 1)δ, respectively. Under such a weighting scheme, if there are a red balls and b
black balls, then the next time a ball is selected from the urn, the probability it is red is
WR (a)
a+δ
a+δ
WR (a)+WB (b) = a+δ+b+(i−1)δ = a+b+iδ . Whenever a ball is picked, it is placed back in the urn
with another ball of the same colour. We can ask, if there are initially a red and b black balls,
and we make n selections, what is the probability that d of those selections are red?
To start with, one may calculate the probability of a particular sequence of n outcomes.
If an n-sequence has d reds followed by n − d blacks, then it has probability pn,d,a,b where
a+δ
a+1+δ
a+d−1+δ
...
a + b + iδ a + b + 1 + iδ
a + b + d − 1 + iδ
b + (i − 1)δ b + 1 + (i − 1)δ
b + n − d − 1 + (i − 1)δ
×
...
a + b + d + iδ a + b + d + 1 + iδ
a + b + n − 1 + iδ
Γ(a + d + δ) Γ(b + n − d + (i − 1)δ) Γ(a + b + iδ)
=
.
Γ(a + δ)
Γ(b + (i − 1)δ)
Γ(a + b + n + iδ)
pn,d,a,b =
It is not hard to see that this is the same probability for any n-sequence with d reds and
n − d blacks, regardless of ordering (this is the exchangeability property of the Pólya urn
9
process). As such, letting XR (n, a, b) be the number of reds picked when n selections are
made, we have
n
n Γ(a + d + δ) Γ(b + n − d + (i − 1)δ) Γ(a + b + iδ)
P(XR (n, a, b) = d) =
pn,d,a,b =
.
d
d
Γ(a + δ)
Γ(b + (i − 1)δ)
Γ(a + b + n + iδ)
(2)
.
With
every
vertex
t
=
Now let i ≥ 2 and consider the vertices [1, i] in (PAt (m, δ))∞
t=i
i + 1, i + 2, . . ., there are m edges created, some of which may connect to vertices in [1, i]. We
ask, what is the probability that an edge connects to i, given that it connects to some vertex
in [1, i]? A coupling with the above Pólya urn process is immediate: after the creation of
PAi (m, δ), we create an urn with Di (i) red balls and 2mi − Di (i) black balls. Every time a
vertex t > i connects an edge into the interval [1, i], a selection is made in the urn process. A
red ball is chosen if and only if the edge connects to i.
To demonstrate that the probabilitiesPcorrespond, suppose in PAt,j−1 (m, δ) we have
i−1
Dk (t, j − 1), suppose also Si−1 (t, j − 1) = b.
Di (t, j − 1) = a. Denoting Si−1 (t, j − 1) = k=1
a+δ
Then it is easily checked that P (g(t, j) = i | g(t, j) ∈ [1, i]) = a+b+iδ
. Hence, if in PAt (m, δ)
there are n edges with one end in [1, i] and the other end in [i + 1, t], then the probability
that d of those edges are attached to vertex i is given by (2). As such, we have the following
proposition.
Proposition 7. Let m ≥ 1, i ≥ 2 be integers and let δ > −m be a real number. Suppose a
Pólya urn process starts with m ≤ a ≤ 2m red and b = 2mi − a black balls, and has weighting
functions WR (k) = k + δ and WB (k) = k + (i − 1)δ for the red and black balls, respectively.
Let the random variable XR (n, a) = XR (n, a, 2mi − a) count the total number of red choices
after n selections have been made. Furthermore, consider a random graph PAt (m, δ). If t ≥ i,
then for 0 ≤ d ≤ n,
P (Di (t) = d + a | Si (t) − 2mi = n, Di (i) = a) = P(XR (n, a) = d).
The following lemma will be used to bound individual vertex degrees.
Lemma 8. Let XR (n, a) be the random variable defined in Proposition 7 and let I = i(2m +
δ) − 1. Then for 1 ≤ d ≤ n,
(m,δ)
1
P(XR (n, a) = d) .
d
Id
I +n−d
and
P(XR (n, a) = 0) .
I
I +n
a+δ
dI
e− I+n ,
a+δ
.
Proof. As per Equation (2),
n Γ(a + d + δ) Γ(b + n − d + (i − 1)δ) Γ(a + b + iδ)
P(XR (n, a) = d) =
.
d
Γ(a + δ)
Γ(b + (i − 1)δ)
Γ(a + b + n + iδ)
10
(3)
(4)
That is, since a + b = 2mi and a + b + iδ = i(2m + δ), we have
n Γ(a + δ + d) Γ(i(2m + δ) + n − (a + δ + d)) Γ(i(2m + δ))
P(XR (n, a) = d) =
d
Γ(a + δ)
Γ(i(2m + δ) − (a + δ))
Γ(i(2m + δ) + n)
We re-write the above as
n Γ(a + δ + d) Γ(I + 1 + n − (a + δ + d)) Γ(I + 1)
P(XR (n, a) = d) =
.
d
Γ(a + δ)
Γ(I + 1 − (a + δ))
Γ(I + 1 + n)
(5)
Suppose first that d > 0. We can write the above as
P(XR (n, a) = d) =
Γ(I + 1)
(n)d Γ(I + 1 + n − (a + δ + d))
Γ(a + δ + d)
d!Γ(a + δ) Γ(I + 1 − (a + δ))
Γ(I + 1 + n)
(6)
((n)d denotes the falling factorial (n)d = n(n − 1) . . . (n − d + 1)).
To bound the above, we shall use the following fact: For real x > 0,
√
1
Γ(x + 1) = cx 2πe−x xx+ 2
1
where cx ∈ [1, e 12x ]. Suppose now x → ∞ and a is a constant. Then, the above implies that
when x + a > 0,
Γ(x + a)
= xa (1 + O(1/x)).
(7)
Γ(x)
Now we bound (6): using (7), we deduce that
Γ(a + δ + d) (m,δ) a+δ−1
. d
.
d!Γ(a + δ)
Also by (7),
Now,
Γ(I+1))
Γ(I+1−(a+δ))
(m,δ)
. I a+δ , and so
(m,δ) 1
Γ(a + δ + d)
Γ(I + 1)
.
(Id)a+δ .
d!Γ(a + δ) Γ(I + 1 − (a + δ))
d
(8)
(n)d Γ(I + 1 + n − (a + δ + d))
n
n−1
n − (d − 1) Γ(I + 1 + n − (a + δ + d))
=
...
.
Γ(I + 1 + n)
I +nI +n−1
I + n − (d − 1)
Γ(I + n − (d − 1))
We have
n
n−1
n − (d − 1)
...
≤
I +nI +n−1
I + n − (d − 1)
n
I +n
d
Furthermore, by (7)
Γ(I + 1 + n − (a + δ + d)) (m,δ)
1
.
.
Γ(I + n − (d − 1))
(I + n − d)a+δ
Consequently, we have the following bound:
11
dI
≤ e− I+n .
(m,δ)
1
P(XR (n, a) = d) .
d
Id
I +n−d
a+δ
dI
e− I+n .
Now suppose d = 0, then going back to (5) we have
Γ(I + 1 + n − (a + δ)) (m,δ)
Γ(I + 1)
.
P(XR (n, a) = 0) =
Γ(I + 1 − (a + δ))
Γ(I + 1 + n)
4.2
I
I +n
a+δ
.
Proof of Theorem 1
For convenience, we rewrite as a(t) = ω 10 ac (t) where ω = ω(t) → ∞ arbitrarily slowly (we
can assume ω ≤ log t, since if not, we can just substitute log t for it and get full infection
whp; a larger ω can only increase the probability of this happening).
Let κ = dω 1+δ/m e and choose [κ] as a core. We wish to show all vertices in the core are
infected for this a(t).
For δ ≥ 0, we apply Lemma 5, taking h to be a sufficiently small constant such that for
some constant K` , we have Sκ (t) ≥ K` tγ κ1−γ whp. For δ < 0, we apply Lemma 6 to get the
same result. We set n = nκ (t) = K` tγ κ1−γ − 2mκ. γ
t
1
Now we wish to show that whp, Di (t) ≥ ω1+δ/m
z over all i ∈ [κ], for some appropriately chosen z = z(t) → ∞. Applying Lemma 8 with I = κ(2m + δ) − 1,
n
P XR (n, a) ≤
=
κz
n/(κz)
X
P(XR (n, a) = d)
d=0
.
≤
I
I +n
a+δ
I
I +n
a+δ
n/(κz) +
X
d=1
+
I
I +n−d
X
a+δ−1
d
Z
d=0
n/(κz)
xa+δ−1 dx ≤
.
0
d=0
dI
da+δ−1 e− I+n
n/(κz)
X
I a+δ
da+δ−1 .
a+δ
(I + n − n/(κz))
Since κ → ∞ and z → ∞ as t → ∞, we have n/(κz) = o(n), so
Furthermore,
n/(κz)
a+δ
1
(I+n−n/(κz))a+δ
1
.
(I+n)a+δ
.
1 n a+δ
.
a + δ κz
Choosing z = ω 2 , we have
I
I +n
a+δ
=
κ(2m + δ) − 1
κ(2m + δ) − 1 + K` tγ κ1−γ − 2mκ
a+δ
.
ω 1+δ/m
t
!γ(a+δ)
a+δ
=o
1
.
z a+δ
Hence,
n
P XR (n, a) ≤
.
κz
I
I +n
a+δ
+
I
I +n
a+δ 12
n a+δ
.
κz
I
I +n
+
1
z a+δ
≤
1
z m+δ
.
Thus,
n
1
P XR (n, a) ≤
. m+δ .
κz
z
Taking a union bound over all vertices in [κ], we have a probability asymptotically bounded
1+δ/m
by zωm
= o(1).
γ
1
t
for each i ∈ [κ], we calculate the expectation of the number
So given Di (t) ≥ ω1+δ/m
ω2
of infected neighbours a vertex in the core has. This would be at least
γ
γ
t
1
a(t)
1
ω8
=
≥ ω7
2mt ω 1+δ/m
ω2
2m ω 1+δ/m
for large enough t.
To calculate the probability that at least r neighbours are infected for a fixed vertex i in
the core, we bound the corresponding binomial random variable. Suppose N = N (t) → ∞,
p = p(t) → 0 but N p → ∞. Then for large enough t, P(Bin(N, p) < r) ≤ e−N p/2 .
Therefore,
γ
a(t)
t
1
7
P Bin Di (t),
< r | Di (t) ≥
≤ e−ω /2
2
1+δ/m
t
ω
ω
7
and so the probability that any of the core vertices fail to be infected is at most ω 1+δ/m e−ω /2 ≤
6
e−ω , for large enough t.
Thus, at this stage, we have proved that the core vertices, i.e., those in [κ], all get infected
whp. If no vertex outside the core has more than a single self-loop, then each vertex will
have at least m − 1 forward (i.e., out-going) edges. Hence, if r ≤ m − 1, the entire graph
will be infected if the core is. We show that no vertex outside the core has more than one
self-loop.
−2
The probability that vertex i outside thecore has at least
two self loops is at most 2 m
2 i .
Rt
Summing over all i ∈ [κ + 1, t], this is O ω1+δ/m i−2 di = O 1/ω 1+δ/m = o(1). Hence,
whp, no vertex outside the core has more than one self-loop. So if r ≤ m − 1, the entire
graph gets infected whp.
5
Subcritical Case: Proof of Theorem 2
The general proof strategy of Theorem 2 is based on the following argument. Suppose that
a vertex i is not infected at round τ = 0, but it is infected at round τ = 1. Then there must
be r edges connected to i that also connect to vertices infected in round τ = 0. Assuming
that these edges connect to different neighbours, we have a depth-1 tree. Similarly, if i gets
infected in round τ = d, then there must be some underlying witness structure which caused
this. In particular, it may be that there is an r-ary tree of depth d wherein in round τ = 0
all the leaves are infected and no internal vertices are. We call this a witness tree. More
generally, such a structure may contain cycles. We shall deal with witness trees first before
addressing more general witness structures. We use a first moment argument to show that
witness structures of a certain depth do not exist whp. Before doing so, we need to develop
the estimates that will allow us to bound the number of occurrences of a certain graph as a
subgraph of PAt .
13
We revert to the model PAt (1, δ), which, for notational convenience, we shall write as PAt .
1
We have γ = 2+δ
.
We begin by defining a sequence of polynomials (Qn (x))n≥1 where Qn (x) = x(x+1) · · · (x+
n − 1).
Lemma 9. Let Xt be a random variable measurable with respect to PAt . Then for 1 ≤ i ≤ t−1
E[Xt−1 Qn (Di (t) + δ)] = E[Xt−1 Qn (Di (t − 1) + δ)]
t + (n − 1)γ
.
t−γ
Proof.
E[Xt−1 Qn (Di (t) + δ) | PAt−1 ]
Di (t − 1) + δ
Di (t − 1) + δ
= Xt−1 1 −
Qn (Di (t − 1) + δ) +
Qn (Di (t − 1) + δ + 1)
(2 + δ)t − 1
(2 + δ)t − 1
Di (t − 1) + δ Di (t − 1) + δ + n
= Xt−1 Qn (Di (t − 1) + δ) 1 −
+
.
(2 + δ)t − 1
(2 + δ)t − 1
Now we take expectations on both sides and the lemma follows.
Lemma 10. Suppose i < j1 , j2 , . . . , jk are vertices in PAt (m, δ). Then
P (j1 → i ∩ j2 → i ∩ . . . ∩ jk → i) ≤ M k
1
1
iγ j11−γ iγ j21−γ
...
1
iγ jk1−γ
where M = M (m, δ) is a constant that depends only on m and δ.
Proof. We begin with the case m = 1. Assuming that j1 < · · · < jk we have
Di (jk − 1) + δ
E[1{j1 →i} 1{j2 →i} . . . 1{jk →i} | PAjk −1 ] = 1{j1 →i} 1{j2 →i} . . . 1{jk−1 →i}
(2 + δ)jk − 1
!
k−1
Y
γ
1{js →i} (Di (jk − 1) + δ).
=
jk − γ
s=1
Therefore, applying Lemma 9 repeatedly,
" k−1
!
#
Y
γ
E[1{j1 →i} 1{j2 →i} . . . 1{jk →i} ] =
E
1{js →i} Q1 (Di (jk − 1) + δ)
jk − γ
s=1
" k−1
!
#
Y
γ
jk − 1
=
E
1{js →i} Q1 (Di (jk − 2) + δ)
jk − γ
jk − 1 − γ
s=1
.
= ..
=
γ
E
jk − γ
"
k−1
Y
s=1
14
!
1{js →i}
#
Q1 (Di (jk−1 ) + δ)
jY
k −1
s=jk−1 +1
s
.
s−γ
Now,
k−1
Y
"
E
1{js →i}
s=1
k−2
Y
#
!
(Di (jk−1 ) + δ) | PAjk−1 −1
=
!
Di (jk−1 − 1) + δ
(Di (jk−1 − 1) + δ + 1)
(2 + δ)jk−1 − 1
!
k−2
Y
1{js →i} Q2 (Di (jk−1 − 1) + δ).
1{js →i}
s=1
=
γ
jk−1 − γ
s=1
Thus, by repeated application of Lemma 9,
E[1{j1 →i} 1{j2 →i} . . . 1{jk →i} ]
" k−2
!
# j −1
k
Y
Y
γ
γ
s
=
E
1{js →i} Q2 (Di (jk−1 − 1) + δ)
jk − γ jk−1 − γ
s−γ
s=1
s=jk−1 +1
" k−2
!
# jk−1 −1
k −1
Y
Y s + γ jY
γ
γ
=
E
1{js →i} Q2 (Di (jk−2 ) + δ)
jk − γ jk−1 − γ
s−γ
s=1
s=jk−2 +1
s=jk−1 +1
s
.
s−γ
This pattern continues until we get
E[1{j1 →i} 1{j2 →i} . . . 1{jk →i} ]
γ
γ
γ
...
=
jk − γ jk−1 − γ
j1 − γ
jY
2 −1
s=j1 +1
s + (k − 2)γ
...
s−γ
jk−2 −1
Y
s=jk−3 +1
s + 2γ
s−γ
jk−1 −1
Y
s=jk−2 +1
s+γ
s−γ
jY
k −1
s=jk−1 +1
× E[Qk (Di (j1 − 1) + δ)].
Applying Lemma 9 repeatedly,
E[Qk (Di (j1 − 1) + δ)] = E[Qk (Di (i) + δ)]
jY
1 −1
s=i+1
s + (k − 1)γ
,
s−γ
and
1+δ
1+δ
Qk (1 + δ) +
Qk (1 + δ + 1)
(2 + δ)i − 1
(2 + δ)i − 1
i + (k − 1)γ
= Qk (1 + δ)
.
i−γ
E[Qk (Di (i) + δ)] =
1−
Thus,
E[1{j1 →i} 1{j2 →i} . . . 1{jk →i} ]
=
jY
1 −1
s=i
s + (k − 1)γ
s−γ
jY
2 −1
s=j1 +1
s + (k − 2)γ
...
s−γ
jk−2 −1
Y
s=jk−3 +1
γ
γ
γ
×
...
Qk (1 + δ).
jk − γ jk−1 − γ
j1 − γ
s + 2γ
s−γ
jk−1 −1
Y
s=jk−2 +1
s+γ
s−γ
jY
k −1
s=jk−1 +1
(9)
Observe that
jY
k −1
s=jk−1 +1
s
s−γ
s
Γ(jk )
Γ(jk−1 + 1 − γ)
Γ(jk ) Γ(jk−1 + 1 − γ)
=
=
,
s−γ
Γ(jk−1 + 1)
Γ(jk − γ)
Γ(jk − γ) Γ(jk−1 + 1)
15
s
s−γ
and similarly with the other product terms. Thus, the product will give us
Γ(jk ) Γ(jk−1 + 1 − γ) Γ(jk−1 + γ) Γ(jk−2 + 1 − γ) Γ(jk−2 + 2γ) Γ(jk−3 + 1 − γ)
...
Γ(jk − γ) Γ(jk−1 + 1) Γ(jk−1 − γ) Γ(jk−2 + 1 + γ) Γ(jk−2 − γ) Γ(jk−3 + 1 + 2γ)
Γ(j1 + 1 − γ)
Γ(j1 + (k − 1)γ)
Γ(i − γ)
Γ(j2 + (k − 2)γ)
...
.
Γ(j2 − γ)
Γ(j1 + 1 + (k − 2)γ)
Γ(j1 − γ)
Γ(i + (k − 1)γ)
Also observe that
Γ(jk−1 + 1 − γ) Γ(jk−1 + γ)
jk−1 − γ Γ(jk−1 + γ)
Γ(jk−1 + γ)
=
<
.
Γ(jk−1 + 1) Γ(jk−1 − γ)
jk−1
Γ(jk−1 )
Γ(jk−1 )
A similar argument bounds the other fraction pairs, thereby giving an upper bound on
the product of
Γ(jk ) Γ(jk−1 + γ) Γ(jk−2 + 2γ)
Γ(j1 + (k − 1)γ)
Γ(i − γ)
...
.
Γ(jk − γ) Γ(jk−1 ) Γ(jk−2 + γ)
Γ(j1 + (k − 2)γ) Γ(i + (k − 1)γ)
(10)
For some constant c which depends only on γ, we have Γ(x + γ)/Γ(x) ≤ cxγ . Therefore,
(j ,j
...j1 )γ
.
as the last term is less than 1, (10) is bounded by ck k k−1
ikγ
Going back to (9), we have
E[1{j1 →i} 1{j2 →i} . . . 1{jk →i} ] ≤ Qk (1 + δ)ck
1
1
iγ j11−γ iγ j21−γ
...
1
iγ jk1−γ
(11)
where c is a constant that depends only on γ.
We wish to extend the above result to PAt (m, δ) with m > 1, that is, we wish to bound the
probability of the js connecting to the same vertex i. The js do not have to be distinct. Recall
that vertex i in PAt (m, δ) is created from grouping m consecutive vertices in PAmt (1, δ/m)
and contracting them into one vertex (possibly creating loops and/or parallel edges in doing
so).
Let I = {m(i − 1) + 1, m(i − 1) + 2, . . . , mi} be the set of vertices in PAmt (1, δ/m) that
group to become i in PAt (m, δ). Similarly, we have sets J1 , J2 , . . . , Jk for the js.
Recall that g(j, `) = i means that the `th edge of vertex j is connected to vertex i < j.
(i) (i)
For a vertex i, an integer Ni and a collection of pairs of indices {(jn , `n )}n=1,...,Ni let
Ei =
Ni
\
{g(jn(i) , `(i)
n ) = i}.
n=1
In other words, Ei is the event that the given set of edges are thrown to vertex i. Since
PAt (m, δ) is created from PAmt (1, δ/m), Inequality (11) implies that for some c1 = c1 (m, δ) >
0 we have
1
i
P (Ei ) ≤ cN
(12)
1−γ .
1
Q
(i)
Ni
N
γ
i
i
n=1 jn
The following lemma from [25] states that Ei and Ei0 are negatively correlated for i 6= i0 .
That is, edges thrown to different vertices are negatively correlated.
Lemma 11 (Lemma 11.13 [25]). For distinct vertices i1 , i2 , . . . , i` in PAt (m, δ),
!
`
`
\
Y
Eis ≤
P(Eis ).
P
s=1
s=1
16
The event j1 → i in PAt (m, δ) occurs in PAmt (1, δ/m) when a vertex in J1 throws an
edge to a vertex in I. This can happen in m2 different ways. More generally, the event
j1 → i ∩ j2 → i ∩ . . . ∩ jk → i in PAt (m, δ) can happen in at most m2k different ways. Any
such way corresponds to the realisation of a collection of events Ei1 , . . . , Ei` in PAmt (1, δ/m),
for some ` ≤ m ∧ k. So the above lemma together with (12) imply (11) (with a different
multiplicative constant) for m > 1.
The following is a corollary of Lemmas 10 and 11. The js need not be distinct, and some
of the js may also be is.
Corollary 12. Suppose i1 , j1 , i2 , j2 , . . . , ik , jk are vertices in PAt (m, δ) where is < js for
s = 1, 2, . . . , k. Then
P(j1 → i1 ∩ j2 → i2 , . . . , jk → ik ) ≤ Qk (1 + δ)M k
1
1
iγ1 j11−γ iγ2 j21−γ
...
1
iγk jk1−γ
where M = M (m, δ, k) depends only on m, δ and k.
5.1
Witness structures
In order to show that a vertex i does not get infected in round τ = 1 whp, it suffices to
show that there is no depth-1 witness structure, whp. This can be done by showing that
the expected number of such witness structures is o(1). We shall deal with trees first, where
every internal (non-leaf) vertex has r children.
Let i ∈ PAt (m, δ). A particular tree Ti , rooted at i = root(Ti ) with leaves L = leaves(Ti ),
is a subgraph of PAt (m, δ). If L ⊆ I0 but no other vertex of Ti is in I0 , then Ti is called
a witness tree. For the sake of the analysis, in this section it will be convenient to consider
edges of PAt (m, δ) to be directed, where edge (i, j) is directed from the younger to the older.
Thus, given a Ti the orientations on its edges are already determined and we are not free to
alter them. Suppose vertex j is a child of vertex j 0 in a tree Ti . If j 0 < j, then the edge {j 0 , j}
is directed from j to j 0 and we call (j, j 0 ) an up edge; otherwise we call it a down edge.
→
−
A given tree Ti is a member of a rooted, directed, isomorphism class Ti : this consists
of pairwise isomorphic rooted trees, where the root is labeled by i and the other vertices
have labels in [t] \ {i}. Here, we assume that an isomorphism between members of this class
respects edge orientations.
→
−
Alternatively, we may define T to be a rooted, directed r-ary tree whose vertices are
the variables x0 , x1 , x2 , . . . , xN and x0 is the label/variable of the root. These variables take
→
−
values in [t]. If we set x0 = i, then we denote the resulting tree (or class of trees) by Ti .
→
−
Every assignment of the variables which respects the edge orientations gives rise to a Ti ∈ Ti .
Let d0 = min{d ∈ N : dγ > 1}. As we shall see, we need only consider trees of depth
at most d0 , which is, of course, a constant. Consequently, there is a bounded number of
isomorphism classes, and since each tree is r-ary, no tree has more than rd0 +1 vertices.
We shall deal with the cases rγ > 1 and rγ ≤ 1 separately, starting with the former.
There, it suffices to consider only trees of depth 1.
We require the following lemma.
17
Lemma 13. For infection probability p = O(1/tγ ), no vertex in I0 has parallel edges whp.
k
Proof. Let Xt be a random variable that counts the number of vertices j which throw parallel
edges in PAt (m, δ). Then, by Lemma 10
k
E[Xt ]
= O(1)
j
t X
X
1
j=1 i=1
i2γ j 2(1−γ)
= O(1)
t
X
1
j=1
j 2(1−γ)
t
j
Z
−2γ
x
1
O(1) X 1 − j 1−2γ
dx =
.
2γ − 1
j 2(1−γ)
j=1
Rt
k
δ/m
When δ < 0, we have 1 − 2γ = 2+δ/m
< 0. Hence E[Xt ] = O(1) 1 j −2(1−γ) dj = O(t2γ−1 ).
Therefore, the expected number of vertices that are in I0 and throw parallel edges, or throw
parallel edges to vertices in I0 , is O(tγ−1 ) = o(1).
When δ = 0, we have γ = 1/2 so the sum is O((log t)2 ), giving probability O((log t)2 /tγ ) =
o(1).
When δ > 0, we have 0 < γ < 1/2 giving probability O(log t/tγ ) = o(1).
5.2
rγ > 1
In this section we prove Theorem 2(i). We remind that δ < 0 implies rγ > 1.
By Lemma 13, any vertex i infected in round τ = 1 must be infected by a depth-1 witness
tree.
Lemma 14. Suppose p = ωt1γ . For a vertex i ∈ PAt (m, δ), the expected number of depth-1
witness trees rooted at i is O ωr1irγ .
Proof. Let Ti be such a tree with k up edges and r − k down edges. Specifically, say the
up leaves are vertices j1 , j2 , . . . , jk and the down leaves are vertices jk+1 , jk+2 , . . . , jr . If
Ti ⊆ PAt (m, δ) means that Ti is a subgraph of PAt (m, δ), then
P (Ti ⊆ PAt (m, δ)) ≤ M k
= Mk
1
iγk+(1−γ)(r−k)
1
1
iγ j11−γ iγ j21−γ
...
1
iγ jk1−γ
×
1
1
γ
γ
i1−γ jk+1
i1−γ jk+1
...
1
i1−γ jrγ
1
1
.
1−γ
(j1 j2 . . . jk )
(jk+1 jk+1 . . . jr )γ
Therefore, the expected number of trees in the isomorphism class (i.e., those trees isomorphic to Ti , rooted at i and having the same edge orientations) is bounded from above
by
Z t
k Z i
r−k
1
1
−1+γ
−γ
O(1) γk+(1−γ)(r−k)
j
dj
j dj
= O(1) γk+(1−γ)(r−k) tγk i(1−γ)(r−k)
i
i
i
1
γk
t
= O(1)
.
i
The above is therefore maximised when k = r, that is, when all edges to leaves are up. There
are 2r possible edge orientations, hence, multiplying by the probability that all leaves of such
a tree are infected we get a bound of O(1/ω r )i−rγ for the expected number of depth-1 witness
trees rooted at i.
The proof of Theorem 2(i) is a corollary of the above: summing O(1/ω r )i−rγ over all i
from 1 to t, the condition rγ > 1 ensures we get o(1).
18
5.3
rγ ≤ 1
In this section we prove Theorem 2(ii) and (iii). Recall that the condition rγ ≤ 1 implies
that δ ≥ 0 and, therefore, γ ≤ 1/2.
5.3.1
Witness trees
Recall that d0 = min{d ∈ N : dγ > 1}. We shall consider witness trees of depth at most
d0 . Since d0 is a constant and each internal vertex has precisely r children, there is only a
bounded number of isomorphism classes.
In round τ = 0, there are Θ(t1−γ /ω) infected vertices in expectation. We will show that in
expectation there are o(t1−γ /ω) newly infected vertices in each of rounds τ = 1, 2, . . . , d0 − 1,
and o(1) in round d0 . Consequently, the progression of the outbreak stops at or before round
p
d0 whp and, moreover, by Markov’s inequality, it follows that |If |/|I0 | → 1 as t → ∞.
If i gets infected in round τ = d, it must be the case that there is a depth-d witness
structure which causes this infection. We shall bound from above the expected number of
such witness structures for d = 1, 2, . . . , d0 . In this section, we focus on witness structures
that are trees - the general case is treated in the next section.
→
−
For the purposes of the next section, we will consider an extended isomorphism class T
which is a rooted, oriented tree, whose vetrices are variables, taking values in [t], and every
vertex has at most r children. Assuming that the tree has N + 1 vertices, the variables that
are the labels of the vertices are x0 , x1 , . . . , xN , where x0 is the label of the root vertex. When
these variables are assigned values in [t] that are compatible with the direction of the edges
→
−
→
−
of T and the corresponding edges are present in PAt (m, δ), then we have a realisation of T .
→
−
→
−
Thus, we may view T as the set of all such realisations - we write T ∈ T . For any i ∈ [t], we
→
−
→
−
let Ti denote the restriction of T where the root variable x0 has been set to i.
→
−
Let us consider, in particular, the case of an extended isomorphism class Ti , for some
→
−
→ count the number of trees Ti ∈ Ti such that Ti ⊆ PAt (m, δ) and L =
i ∈ [t]. Let X−
Ti
leaves(Ti ) ⊆ I0 . We have
h
i
→ =
E X−
T
i
1
ωtγ
|L| X
−
→
T i ∈ Ti
P (Ti ⊆ PAt (m, δ)) .
(13)
Since each tree has at most rd0 +1 edges, by Corollary 12 we have
X
−
→
Ti ∈ Ti
P (Ti ⊆ PAt (m, δ)) ≤ C1 (m, δ, r)
X
Y
−
→
Ti ∈ Ti (a,b)∈E(Ti )
1
.
bγ a1−γ
(14)
where C1 (m, δ, r) is some constant that depends only on m, δ, r and E(Ti ) is the edge set of
Ti . (Recall that each edge is oriented from the larger vertex to the smaller one.)
→
−
Since each tree Ti is an assignment of the variables x1 , x2 , . . . , xN of the tree Ti (recall
that x0 = i), to calculate the sum (13) we can perform a sum over all valid assignments. Our
aim is to bound from above the above sum.
To this end, we consider a more general setting in which each vertex xa is associated
with a valuation function va : [t] → R+ . When the variables/vertices xa assume some
value, i.e., some assignment of a vertex in [t], then the corresponding vertices get the value
va (xa ). We consider valuation functions of a certain form. Namely, if xa is not a leaf, then
19
va (j) = (log j)ρa /j ea , where ρa is a non-negative integer and ea is a non-negative real number
such that ea = Aγ + B(1 − γ) with A, B being non-negative integers that satisfy
c(xa ) + A + B ≥ r. Property (A),
where c(xa ) denotes the number of children of xa . Furthermore, if ea = 0, then ρa = 0, and
we call the valuation function trivial. Hence, if xa is an internal vertex with a trivial valuation
function, then it has exactly r children. Now, if xa is a leaf, then either va (j) = ωt1γ = p
(for r = 2 we take ω = log t) or va (j) = (log j)ρa /j ea , where ρa is a non-negative integer and
ea > 0 satisfies Property (A). In the former case, we call the leaf original ; otherwise, we call
it a contraction leaf. The purpose of having a valuation function of this form will become
apparent in the next section, where we consider general witness structures that are not trees.
In those cases we perform a series of operations that convert a general witness structure
into a tree. During these operations, we perform contractions of subtrees (hence the term
contraction leaf ). Effectively, the valuation function is (up to multiplicative constants) the
probability that the vertex/root of the contracted subtree is infected through this subtree.
When a leaf is original, it is meant to be externally infected, whereas a contraction leaf is
infected through a certain sub-tree (that had been) rooted at it. If all the children of a vertex
are original leaves, then it has exactly r children.
→
−
For a vertex xa of T , we define the function fa : [t] → R≥0 recursively: If xa is a leaf,
then fa (j) = va (j). Otherwise, with xa1 , xa2 , . . . , xak being the child variables of xa , where
k ≤ r, with xa1 , . . . xak1 having up edges with xa and the rest down, we set

fa (j) := va (j) 
k1 X
t
Y
s=1
j 0 =j+1


j−1
k
Y
X
1
1
fas (j 0 ) γ 01−γ  
fas (j 0 ) 0γ 1−γ  .
j j
j j
0
(15)
s=k1 +1 j =1
We call fa the weight function of the sub-tree that is rooted at xa .
→
−
It is not hard to see that if the sub-tree T a that is rooted at xa has N (non-root) vertices
xa1 , . . . , xaN , then fa (j) is
fa (j) = va (j)
N
X Y
Y
va` (j` )
j1 ,...,jN `=1
(xa` ,xb`
1
2
1
γ 1−γ 1{j`1 >j`2 } .
−
→ j`2 j`1
)∈E( T a )
However, if the valuation functions of the internal vertices are trivial and all leaves are
original, then simply
Y
1
1 |L| X
f0 (i) ≥
.
γ
γ
ωt
b a1−γ
−
→
Ti ∈ Ti (a,b)∈E(Ti )
Thus by (13) and (14), if we show that
t
X
f0 (i) = o(1),
(16)
i=1
this will imply that whp there are no vertices which are infected through a tree that is
→
−
isomorphic to T .
20
To this end, we will first provide an upper bound on f0 (cf. Lemma 16 and Corollary 18
below). In fact, we will provide a more general upper bound that is applicable to a general
configuration of witness trees. This general form will be useful in the next section where we
analyse the expected number of occurrences of general witness structures.
→
−
→
−
→
−
→
−
Let T (a) denote the subtree of T rooted at xa . Thus, in particular, T (0) = T . Let ρ(a)
→
−
be the number of down edges in T (a) and let `(a) be the number of its original leaves. We
→
−
→
−
→
−
shall also be writing a0 ∈ T (a) to denote that xa0 is a vertex of T (a). For each a0 ∈ T , we
denote by ea0 and ρa0 the exponents of the valuation function va0 of xa0 , as described above.
First we require the following lemma, the proof of which is in the appendix:
Lemma 15. Assume that xa0 is a leaf and let fa0 (j 0 ) = (log j 0 )ρa0 /j 0ea0 , if it is a contraction
leaf, where ea0 > 0 satisfies Property (A). Then for any 1 ≤ j < t we have
(
t
1 1
X
if xa0 is an original leaf
1
γ,
0
,
fa0 (j ) γ 01−γ . ω j ρ −e
a0 ,
a0 j
j j
0 is a contraction leaf
if
x
(log
j)
a
0
j =j+1
and
j−1
X
j 0 =1
1
fa0 (j ) 0γ 1−γ
j j
0

1 1


 ω jγ ,
. (log j)ρa0 j −ea0 ,


 (log j)ρa0 +1 ,
j 1−γ
if xa0 is an original leaf
if xa0 is a contraction leaf and 1 − γ > ea0 .
otherwise
Remark The above lemma effectively shows that it is enough to consider only the case
where original leaves are connected to their parent through an up edge, and so, we shall
henceforth assume this to be the case. We also remind that if all the children of a vertex
are original leaves, then it has exactly r children.
→
−
→
−
Lemma 16. Suppose that T is as above. Let xa be an internal vertex of T or a contraction
leaf. Then uniformly for all j ∈ [t],
fa (j) .
1
ω `(a)
0
(1 ∨ (log j)ρ(a) )
,
j ya
(17)
where ya = Aγ + B(1 − γ) with
A, B satisfying
A + B ≥ r. Furthermore,
P non-negative integers
P
0 = ρ(a) +
→
−
→
0
B > 0 implies ya ≥ `(a)γ + a0 ∈−
e
,
and
ρ(a)
ρ 0.
T (a) a
a0 ∈ T (a) a
Remark The hidden constant factor in (17) depends only on m, δ and r.
Proof. If xa is a contraction leaf, then fa = va by definition, and the lemma is immediate.
Hence, we assume below that xa is an internal vertex.
For the sake of notational convenience we shall omit the ‘1∨’ component of terms of the
form (1 ∨ (log j)ρ ). We shall give a proof by induction starting from the bottom and going
“up” the tree. The reader is reminded that we are ultimately aiming to bound the right-hand
side of (15)
Assume that xa = j and suppose xa1 is a child of xa . We shall consider a number of cases
regarding xa1 and combine them in the end. Consider first of all that xa1 is a leaf. Recalling
21
that for a leaf xa1 the weight function fa1 is defined to be equal to the valuation function va1 ,
we apply Lemma 15. Thus, if xa1 connects to xa by an up edge, we have
(
t
1 1
X
if xa0 is an original leaf
1
γ,
0
fa0 (j ) γ 01−γ . ω j ρ −e
,
a0 j
a0 ,
j j
0 is a contraction leaf
(log
j)
if
x
a
0
j =j+1
and if xa1 connects to xa by a down edge, we have
j−1
X
j 0 =1
1
fa0 (j 0 ) 0γ 1−γ .
j j
(
(log j)ρa0 j −ea0 , if xa0 is a contraction leaf and 1 − γ > ea0
(log j)ρa0 +1
,
j 1−γ
otherwise
.
(Recall, as per the remark after the statement of Lemma 15, we have made the assumption
that original leaves are only connected through up edges).
Now suppose xa1 is an internal vertex and a child of xa . Letting `1 = `(a1 ), the number
of original leaves in the subtree rooted at xa1 , we have by the induction hypothesis, fa1 (j1 ) .
(log j1 )ρ1
1 `1
A1 γ+B1 (1−γ) for some appropriate A1 , B1 and ρ1 as in the statement of the lemma. In
ω
j1
particular, these are such that A1 + B1 ≥ r.
Consider now the case where xa1 = j1 is connected by an up edge. We have
t
X
j1 =j+1
fa1 (j1 )
1
j γ j11−γ
`1 X
t
1
(log j1 )ρ1
1
.
γ 1−γ A1 γ+B1 (1−γ)
ω
j1
j1 =j+1 j j1
`1
Z t
1
1
−1+γ−(A1 γ+B1 (1−γ))
.
x
(log x1 )ρ1 dx1 .
ω
jγ j 1
Observe A1 γ + B1 (1 − γ) > γ. Indeed, since δ ≥ 0 we have γ ≤ 1 − γ. Thereby, A1 γ + B1 (1 −
γ) ≥ (A1 + B1 )γ ≥ rγ > γ. Therefore, we can apply the following lemma to bound the above
integral. Its proof can be found in Section 8.2 (appendix).
Lemma 17. Let k ≥ 0 be an integer, let α > 0 be a real number and let
Z t
Ik,α (j) :=
(log x)k x−1−α dx.
j
Then uniformly for j ≥ 1 we have
Ik,α (j) .
(log j ∨ 1)k
.
jα
This gives
t
X
j1 =j+1
fa1 (j1 )
1
j γ j11−γ
`1
1
(log j)ρ1
.
.
ω
j A1 γ+B1 (1−γ)
(18)
Note that due to the fact that we consider trees of bounded degree and depth, terms such
as ρ and Aγ + B(1 − γ) will always be bounded from above and below by constants that
depend only on m, δ and r. Therefore, the constant factor incurred by the above integration
is always bounded by some constant that only depends on these parameters.
22
Observe that (18) is the same (up to multiplicative constants) as the expression for fa1 (j1 )
except that j has replaced j1 . In this sense, we see that an up edge causes the parent vertex to
“reverse inherit” the exponent of the child, in this case, that exponent being A1 γ + B1 (1 − γ).
Now we will consider what happens if it is a down edge, where, by assumption, xa1 is an
internal vertex. We have
`1 X
j−1
j−1
X
1
1
1
(log j1 )ρ1
fa1 (j1 )
.
γ
1−γ
ω
j j 1−γ j A1 γ+B1 (1−γ)
j γ j1
j1 =1
j1 =1 1
1
`1
Z j
ρ
1
(log j)
1
−γ−(A1 γ+B1 (1−γ))
x1
dx1
.
1−γ
ω
j
1
(
1 `1 (log j)ρ1 +1
if 1 − γ − (γA1 + (1 − γ)B1 ) ≤ 0
ω
j 1−γ
.
(19)
ρ1
`
(log
j)
1 1
if 1 − γ − (γA1 + (1 − γ)B1 ) > 0
ω
j A1 γ+B1 (1−γ)
Once again, we emphasise that the integration incurs a constant factor that is bounded
by a constant that depends only on m, r and δ.
Let C(a) denote the set of indices of the children of xa . For any a0 ∈ C(a), let ra0 denote the
exponent of j in the expression of the upper bound of fa0 (j). We let C1 ⊆ C(a) denote the set
of indices of the original leaves among the members of C(a). Also, we let C2 ⊆ C(a) denote the
set of the indices of those children of xa that are not original leaves but are connected to xa
through up edges. Let C20 denote the set of the indices of those children that are not original
leaves, are connected to xa through down edges and 1 − γ > ra0 , for a0 ∈ C20 . Similarly, we
define as C200 the set of the indices of those children that are not original leaves, are connected
to xa through down edges but 1 − γ ≤ ra0 , for a0 ∈ C200 .
By the above and the the definition in (15),
P
Pa0 ∈C(a) `(a0 )
|C200 |+ a0 ∈C ∪C 0 ∪C 00 ρ(a0 )
2
2
2
1
(log j)
P
fa (j) . va (j)
(20)
|C1 |γ+ a0 ∈C ∪C 0 ra0 +(1−γ)|C200 |
ω
2
2
j
P
Let ya denote the exponent of j. Firstly, note that `(a) = a0 ∈C(a) `(a0 ).
Assume that C2 ∪ C20 = ∅. Then ya = |C1 |γ + (1 − γ)|C200 | + ea . But as |C1 | + |C200 | = c(xa )
and ea satisfies Property (A), it follows that |C1 |γ + (1 − γ)|C200 | + ea = Aγ + B(1 − γ), where
A, B are non-negative integers that satisfy A + B ≥ r.
If C2 ∪ C20 6= ∅, then ra0 > 0, for some a0 ∈ C2 ∪ C20 , which has the form Aγ + B(1 − γ),
for some A, B that are non-negative integers satisfying A + B ≥ r. Hence, the exponent of j
satisfies this as well.
Assume now that ya cannot be written in the form Aγ + B(1 − γ) with A, B non-negative
integers andPB > 0. Then this is the case for ra0 for any a0 ∈ C2 ∪ C20 . Hence, by the induction
hypothesis a0 ∈C2 ∪C 0 ra0 is equal to the number of original leaves that are contained in the sub2
tree that is rooted at those xa0 together with the sum of the exponents ea0 of the valuation
functions of the vertices of these sub-trees. Moreover, |C200 | = 0 and recall that |C1 | is the
number of original leaves that are directly connected to xa . Thereby,
X
ya = `(a)γ +
ea0 .
−
→
a0 ∈ T (a)
The base case, where xa has only leaves for children, is immediate from the above.
23
The above lemma now implies the following.
→
−
Corollary 18. If the valuation functions of the internal vertices of T are trivial and all
leaves are original, then
` (1 ∨ (log i)ρ )
1
(21)
f0 (i) .
ω
iy0
where ` = `(0) and ρ = ρ(0) and either y0 = Aγ + B(1 − γ) where A, B are non-negative
integers that satisfy A + B ≥ r and B > 0 or y0 = `γ.
We conclude with the proof of (16) for depth d0 . Consider the expression on the righthand side of (21). If 0 < B < r and r ≥ 3, then A ≥ r − B and it is easy to check that
Aγ + B(1 − γ) > 1 (it is a convex combination of two positive numbers that are at least 1, one
of which is bigger than one, where γ 6= 0, 1). If B ≥ r, then Aγ+B(1−γ) ≥ r(1−γ) ≥ 3(1−γ).
But γ ≤ 1/2, whereby 3(1 − γ) ≥ 3/2 > 1. If y0 = `γ, then ` ≥ d0 implies y0 ≥ γd0 > 1.
If r = 2 then we are not necessarily guaranteed Aγ + B(1 − γ) > 1 since, for example,
B = 1 and r = 2 only assures γA + (1 − γ)B ≥ 1.
If γA + (1 − γ)B > 1, then
` X
` Z t
` !
t
t
X
1
(log x)ρ
(1 ∨ (log i)ρ )
1
1
f0 (i) .
.
(22)
.
dx
=
O
γA+(1−γ)B
ω
ω
ω
iγA+(1−γ)B
1 x
i=1
i=1
In this case, the expected number of witness trees of this isomorphism class, over all i, is o(1).
`
In the case that y0 = `γ ≤ 1, the sum is ω1 (log t)ρ+1 t1−`γ and the expected number
of witness trees of this isomorphism class, over all i, is O((log t)ρ+1 t1−`γ /ω ` ) = o(t1−γ /ω ` )
since ` ≥ 2. In other words, the expected number of witness trees of depth less than d0 is
o(t1−γ /ω ` ).
As stated above, there are only a bounded number of isomorphism classes that we need
to consider, hence the relevant constant factors are absorbed into the O(.) terms above.
We would like to extend this to include r = 2, wherein if we take the depth of the tree
to be equal to d0 , then it may be the case that the exponent of i in f0 (i) is 1 (which is the
minimum it can be when ` ≥ d0 ). In that case, the integral in (22) would grow like (log t)ρ+1 .
To bypass this difficulty, when r = 2 we consider witness trees that have depth equal to d0 +1.
Recall that in this case we assume that p0 = log1 t t1γ . Also, as we have already commented in
the remark after Lemma 15, we may assume that the witness trees we consider are such that
all their leaves are connected to the rest of the tree through up edges.
Let x1 , x2 be the children of x0 and assume without loss of generality that the subtree
that is rooted at x1 has depth d0 . Suppose the exponent of j1 is 1. That is, recalling that
ρ(1)
this subtree has `(1) original leaves, by Lemma 16 we have f1 (j1 ) . (log 1t)`(1) (log jj11) . Thus,
if x1 is connected by an up edge with x0 , by (18) the exponent transfers, and we get a factor
(log i)ρ(1)
1
i
(log t)`(1)
in f0 (i). If it is connected through a down edge, by (19), we get the factor
1
1
1−γ
`(1)
i
(log t)
Z
1
i
x−γ−1
(log x1 )ρ(1) dx1 .
1
24
1
1
.
1−γ
`(1)
i
(log t)
If x2 is an original leaf, then by Lemma 15 it contributes a factor that is at most (up to a
1
multiplicative constant) (log1 t) i1γ , thus giving in total (log t)1`(1)+1 i1+γ
or (log t)1`(1)+1 1i . In any
case,
f0 (i) .
1
1
.
(log t)`(1)+1 i
If x2 is not an original leaf, then by Lemma 16 f2 (j2 ) .
(23)
(log j2 )ρ(2)
1
,
y
(log t)`(2)
j2 2
where either y2
can be written as Aγ + B(1 − γ) for some non-negative integers A, B that satisfy B ≥ 1 and
A + B ≥ 2, or y2 ≥ `(2)γ, where in this case `(2) ≥ 2.
If x2 is joined to x0 by an up edge, then by (18) it contributes a factor that is at most
(up to a constant)
f0 (i) .
(log i)ρ(2)
1
iy 2
(log t)`(2)
.
(log i)ρ(2)
1
,
i2γ
(log t)`(2)
giving a total
(log i)ρ(1)
1
(log i)ρ(2)
1
(log i)ρ(0)
1
=
.
i2γ
(log t)`(1) i1−γ (log t)`(2)
(log t)`(0) i1+γ
(24)
If x2 is not an original leaf and is connected to x0 by a down edge, the possibilities are
f0 (i) .
1
1
(log i)ρ(1)
(log i)ρ(2)+1
,
i
i
(log t)`(1)
(log t)`(2)
f0 (i) .
1
(log i)ρ(1)
1
(log i)ρ(2)+1
,
i
i2γ
(log t)`(1)
(log t)`(2)
or
or
f0 (i) .
1
1
1
(log i)ρ(2)+1
,
1−γ
i
(log t)`(1) i
(log t)`(2)
f0 (i) .
1
1
1
(log i)ρ(2)+1
.
i2γ
(log t)`(1) i1−γ (log t)`(2)
or
In all cases,
f0 (i) .
1
(log i)ρ(0)
.
(log t)`(0) i1+γ
(25)
Summing (23), (24) or (25) over i = 1, . . . , t gives o(1).
Consequently, the expected number of witness trees of depth d0 + 1 when the initial
infection probability is p0 = (log1t)tγ is o(1) as well.
We have shown that if r ≥ 3, then whp the process stops in less than d0 rounds, whereas
for r = 2 (with the appropriate choice of p0 ) it stops in at most d0 rounds. Note that
d0 = b γ1 c + 1. To be more precise, we have shown the bounds of Theorem 2 for witness
structures that are trees. We need to argue about general witness structures that may contain
cycles. In this case, we show that the expected number of occurrences of such a structure is
bounded by the expected number of occurrences of a tree that is appropriately constructed
and has depth either d0 or d0 + 1, depending on the value of r.
25
5.3.2
General witness structures
We now consider general witness structures that may have cycles. Recall that we are only
assuming that δ ≥ 0, since δ < 0 ⇒ rγ > 1. Recall that in this case γ ≤ 1/2.
Firstly, the following lemma allows us to consider witness structures where the initially
infected vertices are vertices which do not belong to cycles.
Lemma 19. Let K be positive constant. If p = O(1/tγ ), then with high probability, no
initially infected vertex lies on a cycle of size at most K
Proof. For a cycle C = (a1 , a2 , . . . , ak ) of size k ≤ K, we apply Corollary 12: for some
constant M = M (K, δ, m) we have
P(C ⊆ PAt (m, δ)) ≤ M 2k
k
Y
i=1
1
M 2k
≤
,
(ai ∧ ai+1 )γ (ai ∨ ai+1 )1−γ
a1 . . . ak
where we have used the fact that for i < j, iγ j11−γ ≤ (ij)11/2 when γ ≤ 21 .
Thus, the expected number of cycles in PAt (m, δ) of size at most K is bounded from above
by
X X
M 2k
= O((log t)K )
(26)
a
.
.
.
a
1
k
a ,...,a
3≤k≤K
1
k
and so the number of initially infected vertices on such cycles is O((log t)K+1 /tγ ) = o(1).
However, it is not true that initially infected vertices lie far from cycles. Hence, when one
is supposed to trace the infection history, one has to take into account those local structures
that are not trees. Furthermore, a vertex may get its first infection through such a structure.
We will now formalise this notion.
Recall that if a vertex i becomes infected in round τ , then it must have been infected by
some neighbours, at least one of which got infected in round τ − 1. Iterating this argument,
there must be a chain of infections of length τ that started in a set of initially infected vertices.
This is witnessed by a rooted subgraph, whose root is vertex i and whose other vertices can
be classified according to their depths. Let us consider this notion more precisely. Suppose x
and y are neighbours in PAt (m, δ), x ∈ I(τ ) ∩ S(τ − 1), and y ∈ I(τ − 1). Then we say x is
a parent of y and y a child of x. If x is a parent of y and x < y then {x, y} is an up edge. If
x > y, then it is a down edge. The notion of parent-child gives rise to the depth of a vertex.
Let depth(i) = 0 and depth(y) = 1 + max{depth(x) : y is a child of x}. We shall use this
notion later in our proof.
Suppose a vertex i ∈ I(τ ) for some τ > 0. Then there must exist a subgraph Si ⊆
PAt (m, δ) such that the following hold:
(1) every vertex in Si except i has a parent in Si ;
(2) the set L = Si ∩ I(0), which we call the set of leaves, is non-empty;
(3) every parent in Si has exactly r edges in Si which go to children in Si .
If, furthermore, i ∈ I(τ ) ∩ S(τ − 1), that is, i got infected in round τ , then we also have:
26
(4) depth(Si ) = maxj∈Si depth(j) = τ , where depth(i) = 0.
We call such an Si a witness structure rooted at i. Observe that (1) forces Si to be
connected. Of course, leaves cannot be parents. Additionally, recall that we only need to
analyse bounded size structures and, therefore, only have bounded size cycles. Hence, by
Lemmas 13 and 19, any leaf will, with high probability, have degree 1 in Si . We will assume
this to be the case.
Condition (3) implies that a parent has at most r children in Si , and the witness structure
is a witness tree as per the previous definition, if and only if every parent has exactly r children
and every vertex except i has exactly one parent. For a tree, it is also the case that the depth
as defined here in terms of infections coincides with the standard meaning of depth – the
graph distance from the root i to a vertex.
Our aim is to bound from above the expected number of witness structures that are
rooted at i. To this end, we will bound this expected value by the expected number of
occurrences of a tree which is produced from this witness structure through a bounded number
of transformations. Informally, during each transformation we “destroy” vertices which belong
to cycles in this witness structure. Eventually, having destroyed all such vertices we will obtain
a tree whose vertices are equipped with certain valuation functions. We finally bound the
expected number of occurrences of this tree using Lemma 16.
→
−
As with trees, we let S denote an isomorphism class of a witness structure. This can be
viewed as a directed graph whose vertices x0 , . . . , xN are variables taking values in [t], that
satisfies Conditions (1) and (3). We assume that its root is x0 . If S is a witness structure
→
−
on [t] that is isomorphic to S , where adjacent vertices are compatible with the directions
→
−
→
−
→
−
of the corresponding edges of S , then we write S ∈ S . We let Si denote the subset of the
→
−
isomorphism class S , where the root is vertex i. That is, x0 = i.
→
−
→ count the number of copies Si ∈ Si such that Si ⊆ PAt (m, δ) and L = leaves(Si ) ⊆
Let X−
Si
I0 . We have
h
i 1 |L| X
→ =
E X−
P (Si ⊆ PAt (m, δ)) .
(27)
Si
ωtγ
−
→
Si ∈Si
Using Corollary 12 we have
X
−
→
Si ∈Si
P (Si ⊆ PAt (m, δ)) ≤ C2 (m, δ, r)
X
Y
−
→
Si ∈Si (a,b)∈E(Si )
1
bγ a1−γ
.
where C2 (m, δ, r) is some constant that depends only on m, δ, r and E(Si ) denotes the edge
set of Si .
As in the case of trees, we will consider the notion of a generalised witness structure, where
each vertex xa is associated with a valuation function va : [t] → R+ . The valuation functions
we consider are as those we considered in the previous section.
→
−
→ : [t] → R+ , which generGiven such a witness structure S , we will define a function f−
S
alises the weight function of a tree that was defined in the previous sub-section. When the
→ (i) is (up to multiplicative constants) the expected
valuation functions are trivial, then f−
S
→
−
number of occurrences of S rooted at i, in the product space of PAt (m, δ) and the set of
→
−
initially infected vertices. Assume that the vertices of S are x0 , , . . . , xN , where x0 is the
root. We will be associating the index ja with the variable xa . Also, recall that the edges of
27
→
−
S are directed and therefore the edges are ordered pairs. Letting j0 = i, we set
→ (i) = v0 (i)
f−
S
N
X Y
va (ja )
j1 ,...,jN a=1
Y
1
γ 1−γ 1{ja >jb } .
−
→ jb ja
(xa ,xb )∈E(Si )
This should not be confused with notation introduced in (15). However, it is not hard to see
→
−
that if S is a tree, then the above function coincides with the function f0 (i).
→
−
Fix a directed isomorphism class S . We demonstrate how a sequence of transformations
→
−
→
−
can transform S into a tree isomorphism class T , such that each class in the sequence is
an upper bound (in terms of expectation of witness structures) for the previous. Note that
by Lemmas 13 and 19, it suffices to consider witness structures of bounded depth where all
initially infected vertices have degree 1.
Let xa be a vertex on a cycle such that it has maximum depth (as defined above in terms
→
−
of the parent-child relation) among all vertices on cycles. Let T (a) be the sub-tree rooted at
→
−
vertex xa . We apply Lemma 16 to T (a) and obtain
→ (ja ) .
f−
T (a)
1 (log ja )ρ(a)
,
jaya
ω `(a)
where ρ(a) and ya are as in Lemma 16. In particular, ya = Aγ + B(1 − γ), where A, B are
non-negative integers that satisfy A + B ≥ r.
→
−
We are now ready to define the witness structure T S . Assume that xa has k > 1 parents xa1 , . . . , xak (not necessarily distinct). Also assume that xa is connected to xa1 , . . . , xah
through up edges and to xah+1 , . . . xak through down edges, where 0 ≤ h ≤ k. Let ∆ now be
→
−
the index of a parent of the highest depth among xa1 , . . . , xah , if h > 0. To construct T S
→
−
1. remove T (a) together with xa ;
a
2. multiply va∆ (ja∆ ) by (log ja∆ )ρ(a) ja−y
∆ ;
3. multiply vai (jai ) by ja−γ
i , for all i 6= ∆ and i ≤ h;
(1−γ)∧ya
4. multiply vai (jai ) by (log jai )ρ(a)+1 /jai
, for all i = h + 1, . . . , k.
If one of the xai s is connected to xa through parallel edges, then the appropriate step from
the above is applied once for each edge. For the particular case of xa∆ , Step 2 is applied once
for one of the parallel edges, whereas for the others we apply Step 3. If the parallel edges are
down edges, then we apply Step 4 once for each of them.
Note that if the valuation functions vai which are modified have exponents eai satisfying
Property (A), then the modifications incurred by Steps 2-4 preserve this property. Steps 3 and
4 simply remove a child of xai and add to the exponent eai a γ or a ya ∧ 1 − γ, thus preserving
Property (A). Step 2 removes a child of xa∆ and adds ya to ea∆ . But ya = Aγ + B(1 − γ),
for some non-negative integers A, B that satisfy A + B ≥ r. Hence, Property (A) is also
preserved for this exponent.
Steps 2-4 yield
(
X
X
k min{1 − γ, ya }
if h = 0
ea0 =
ea0 +
.
(28)
(h − 1)γ + ya + (k − h) min{1 − γ, ya } if h > 0
−
→
−
→ −
→
0
0
a ∈T S
a ∈ S \ T (a)
28
As we shall see in the proof of the next lemma, Steps 2-4 essentially correspond to a step
→
−
among a sequence of steps that transform S into a tree. In each step, we have the creation
of copies of xa , which we denote by xa(1) , . . . , xa(k) , where xa(i) is attached to xai through an
up edge if i ≤ h or through an down edge if i > h. Thereafter, xa(∆) as well as xa(i) , for i > h,
→
−
each becomes the root of a copy of T (a) (cf. the remark below), whereas for the remaining
is, the vertices xa(i) become original leaves (cf. Figure 1). We denote the resulting directed
→
−
→
−
graph by T̂ S . Note that this is not the directed graph T S . The latter may be thought as
→
−
coming from T̂ S with the subtrees rooted at each xa(i) contracted into xai , multiplying the
corresponding valuation functions of xai by certain factors, as in Steps 2-4. These factors are
upper bounds on the probability that xa(i) will be infected through the sub-tree that is rooted
at it.
→
−
→
−
Remark Note also that the depth of T̂ S is equal to the depth of S . This is the case as all
→
−
xa(i) , for i > h, are the roots of a copy of T (a) as well as xa(∆) . The latter is adjacent to
the deepest parent xa∆ among the xai s, for i ≤ h. This is the reason for which we “assign” a
→
−
rooted copy of T (a) to this particular vertex and we do not treat it as an original leaf, as we
did in the case of the other vertices that are connected to xa through up-edges.
→
−
Because of (27), we are interested in the case
P where the initial witness structure S has1
→ ea0 = 0. Now make the conservative
only trivial valuation functions. In this case, a0 ∈−
S
assumption that each time we apply T , we have 1 − γ ≥ ya , which means each ya will have the
form Aγ for some integer A ≥ r. It then follows from (28) that during the jth transformation
the sum of the exponents of the valuation functions increases by γ`j , where `j is the number
→
−
→
−
of leaves that are added during the transition from T̂ (j−1) S to T̂ (j) S . This can be seen
inductively: original leaves with up edges contribute γ each and each ya = Aγ represents the
accumulation of γ terms from leaves in the subtree rooted at xa .
→
−
Assume that the process stops after step j0 . Thus, T̂ (j0 ) S is an r-ary tree where all its
leaves are original and, by the above remark, has depth that is equal to the maximum depth
→
−
→
−
in S . If Lj0 is the number of leaves of this tree and `(T (j0 ) S ) is the number of original leaves
→
−
of T (j0 ) S , then
X
→
−
Lj0 γ =
ea0 + `(T (j0 ) S )γ.
(29)
−
→
a0 ∈T (j0 ) S
We will use this fact towards the end of our analysis. We now proceed with our ba→ after the application of a sequence of
sic inductive step which will allow us to bound f−
S
transformations T .
→
−
Lemma 20. Let S be a witness structure and xa be a vertex of maximum depth on which
→
−
we perform the above transformation. If `(a) denotes the number of original leaves in T (a),
then uniformly for all i ∈ [t]
1
→ (i) .
→ (i).
f−
fT −
S
S
`(a)
ω
→
−
Proof. Let xa be a vertex of S of maximum depth as in the statement of the lemma. Also,
→
−
we denote the set of indices of the vertices of T (a) by T (a). Let xa1 , . . . , xak be the parents of
→
−
xa . Denote by P(a) the set of indices of S not in T (a). Note that aj ∈ P(a), for j = 1, . . . , k.
1
The assumption is conservative because a larger exponent can only make the final upper bound smaller.
29
xa1
xah
xa∆
xah+1
xak
xa
→
−
T (a)
xa∆
xa1
xah
→
−
T (a)
xak
xah+1
→
−
T (a)
→
−
T (a)
Figure 1: Transformation T̂
Finally assume that the edges (xa , xa1 ), . . . , (xa , xah ), where 0 ≤ h ≤ k are all up edges and
(xa , xah+1 ), . . . , (xa , xak ) are all down edges.
Now for a set of indices S we define the function
Y
Y
1
fS (ja : a ∈ S) =
va (ja )
γ 1−γ 1{ja >jb } ,
j j
a∈S
(x ,x )∈E(S) b a
a
b
where E(S) denotes the set of directed edges that is induced by S. Using this, we write
→ (i) =
f−
S
jah+1 ∧···∧jak ∧t
X
X
ja0 : a0 ∈P(a)\{a1 ,...,ak } ja1 ,...,jak
X
0
fP(a) (ja0 : a ∈ P(a))
fa (ja )
ja >ja1 ∨···∨jah
h
Y
1
k
Y
1
,
γ 1−γ
j γ j 1−γ
i=1 jai ja
i=h+1 a ai
(30)
→
−
where fa is the weight function of T (a). Note that the index i in the LHS defines the index
a in the RHS.
We will take an upper bound for each case of the definition of T with the use of Lemma 16.
In particular, using Lemma 16 we will obtain an upper bound on
jah+1 ∧···∧jak ∧t
X
ja >ja1 ∨···∨jah
fa (ja )
h
Y
1
k
Y
1
.
γ 1−γ
j γ j 1−γ
i=1 jai ja
i=h+1 a ai
(31)
(If h = 0 then we let ja1 ∨ · · · ∨ jah = 0, and if h = k, then we let jah+1 ∧ · · · ∧ jak ∧ t = t.)
30
Applying Lemma 16 to fa (ja ), we obtain
fa (ja ) .
1 (log ja )ρ(a)
,
jaya
ω `(a)
(32)
where ya is as in Lemma 16.
Assume first that h > 0. Then with ∆ as above, we have
jah+1 ∧···∧jak ∧t h
X
Y
1
fa (ja )
γ 1−γ
j γ j 1−γ
i=1 jai ja
i=h+1 a ai
ja =ja1 ∨···∨jah
1
k
Y
1
jah+1 ∧···∧jak ∧t
1
X
1
1
...
(log ja )ρ(a)
j ya
jaγh ja1−γ jaγ ja1−γ
jaγ ja1−γ
h+1
k
Z ja ∧···∧ja ∧t
h+1
k
1 1
1
1 1
. `(a) γ . . . γ 1−γ . . . 1−γ
(log z)ρ(a) z −(k−h)γ−h(1−γ)−ya dz
ja1
jah jah+1
ω
jak
ja =ja1 ∨···∨jah
Z t
1 1
1
1 1
≤ `(a) γ . . . γ 1−γ . . . 1−γ
(log z)ρ(a) z −(k−h)γ−h(1−γ)−ya dz.
ja1
jah jah+1
ω
jak
ja∆
ω `(a)
ja =ja1 ∨···∨jah
jaγ1 ja1−γ
1
...
.
(33)
But by Lemma 17, we have
Z t
1
1
(log ja∆ )ρ(a)
(log z)ρ(a) z −(k−h)γ−h(1−γ)−ya . γ
γ
ja∆ ja∆
ja∆ ja−1+(k−h)γ+h(1−γ)+ya
∆
(log ja∆ )ρ(a)
=
(k−h)γ+(h−1)(1−γ)+ya
ja∆
≤
(log ja∆ )ρ(a)
.
a
jay∆
Thereby, (33) becomes
jah+1 ∧···∧jak ∧t h
Y
X
k
Y
1
1
1−γ fa (ja ) .
γ 1−γ
jγ j
ja =ja1 ∨···∨jah i=1 ja jai
i=h+1 ai a
1 1
(log ja∆ )ρ(a) 1
1
1
. . . γ 1−γ . . . 1−γ .
.
.
.
γ
ya
γ
`(a)
ja1
ja∆
ja∆+1
jah jah+1
ω
jak
1
(34)
Now, if h = 0, then (31) yields
ja1 ∧···∧jak
X
1
ja =1
jaγ ja1−γ
1
...
1
jaγ ja1−γ
k
(log ja )ρ(a)
jaya
(log ja1 )ρ(a)
.
.
ja1−γ
1











...
(log jak )ρ(a)
ja1−γ
k
Z
jak
z −kγ−ya dz
1
)ρ(a)
)ρ(a)
(log jak
(log ja1
a
...
ja1−kγ−y
1−γ
k
ja1−γ
j
a
1
k
(log jak )ρ(a)+1
(log ja1 )ρ(a)+1
.
.
.
ja1−γ
ja1−γ
1
k
(log jak )ρ(a)+1
(log ja1 )ρ(a)+1
ja1−γ
1
...
ja1−γ
k
if 1 − kγ > ya
if 1 − kγ = ya
if 1 − kγ < ya
(35)
In the first case, we have ya < 1 − γ, since ya < 1 − kγ. Thus the last factor is
1
(k−1)γ+ya
jak
≤
1
1
ya = (1−γ)∧ya .
jak
jak
31
Hence, in any case (35) is bounded by
ja1 ∧···∧jak
X
1
ja =1
jaγ ja1−γ
1
...
1
jaγ ja1−γ
k
k
(log ja )ρ(a) Y (log jai )ρ(a)+1
.
.
(1−γ)∧ya
jaya
ja
i=1
(36)
i
Setting

1

γ


 jai
ρ(a) · 1 ,
v̂ai (jai ) := (log jai )
jayia



(log ja )ρ(a)+1 ·
if i ≤ h and i 6= ∆,
if i = ∆ ,
1
(1−γ)∧ya
i
jai
if i > h,
now (34) and (36) yield
jah+1 ∧···∧jak ∧t h
X
Y
ja >ja1 ∨···∨jah
1
k
Y
1
γ 1−γ
γ 1−γ fa (ja )
j
j
j
a
a
a
i
i ja
i=1
i=h+1
.
1
ω `(a)
Qk
i=1 v̂ai (jai ).
(37)
So, substituting the bound of (37) into (30) we obtain
→ (i) .
f−
S
=
1
ω `(a)
X
ja0 :
a0 ∈P(a)\{a1 ,...,ak } ja1 ,...,jak
1
ω `(a)
X
ja0 :
X
X
a0 ∈P(a)\{a1 ,...,ak } ja1 ,...,jak
k
Y
vai (jai )v̂ai (jai )
i=1
=
fP(a) (ja0 : a0 ∈ P(a))
Y
k
Y
v̂ai (jai )
i=1
va0 (ja0 )×
a0 ∈P(a)\{a1 ,...,ak }
Y
(38)
1
1{j 0 >j 0 }
j γ j 1−γ a b
(xa0 ,xb0 )∈E(P(a)) b0 a0
1
→ (i).
f −
ω `(a) T Si
Q
Note that the upper bounds in Lemma 15 imply that ki=1 v̂ai (jai ) is the bound we would
get if xa is replicated k times into xa(1) , . . . , xa(k) and xa(i) is attached to xai through an up
edge if i ≤ h or through an down edge if i > h. Thereafter, xa(∆) as well as xa(i) , for i > h,
→
−
each becomes the root of a copy of T (a), whereas for the remaining is, the vertices xa(i)
become leaves with valuation functions that are equal to 1/tγ . Note that the latter is ωp0 so essentially these become original leaves.
→
−
→
−
→
−
→
−
Starting with the original witness structure S , we get a sequence of structures T S , T (2) S , . . . , T (j) S
→
−
by applying the transformation T in the following way: If T (j−1) S is a tree, we are done;
→
−
otherwise choose a vertex xa of T (j−1) S such that xa is on a cycle and has maximum depth
→
−
among such vertices in T (j−1) S . Now apply to xa and its parents the transformation T , as
→
−
→
−
appropriate, to get T (j) S . Note that T (j) S has at least one less vertex that lies on a cycle.
In general, the number of vertices lying on cycles reduces by at least one each time we apply
→
−
the transformation. Hence, there exists a j0 ≥ 0 such that T (j0 ) S is a generalised witness
→
−
tree. Moreover, the depth of this tree is no more than the depth of S .
32
If xa1 , . . . , xaj0 denote the vertices that were split in each transformation, the repeated
application of Lemma 20 yields
→ (i) .
f−
S
1
ω
Pj0
j=1
→ (i),
f (j ) −
`(aj ) T 0 S
→
−
→
−
where `(aj ) is the number of original leaves of T (aj ) in T (j−1) S .
→
−
Since T (j0 ) S is a generalised witness tree, we can apply Lemma 16 and deduce that for
→
−
some ρ ≥ 0 and with ` being the number of original leaves in T (j0 ) S we have
→ (i) .
f−
S
1
ω
Pj0
`+ j=1
`(aj )
(log i)ρ
,
iy
and either y can be expressed as Aγ + B(1 − γ) where A, B are non-negative integers such
that A + B ≥ r and B > 0 or
X
y=
ea0 + `γ.
−
→
a0 ∈T (j0 ) S
Note that ` +
Pj0
j=1 `(aj )
≥ r.
→
−
By (29) the latter is equal to Lj0 γ, where Lj0 is the number of leaves of T̂ (j0 ) S . But
→
−
T̂ (j0 ) S is an r-ary tree of depth d0 and therefore Lj0 ≥ d0 . Thus, y ≥ γd0 > 1. Hence (22)
also holds in this case, implying that the right-hand side of (27) is o(1). Now, if the depth
of this particular witness structure is less than d0 , then by the same principles as in the case
of trees we obtain an upper bound on the number of vertices that can be infected through a
→
−
witness structure isomorphic to S which is o(t1−γ ).
This completes the proof of Theorem 2(ii) and 2(iii).
6
Critical case
Proof of Theorem 3(i). Let G be a realisation of PAt (m, δ). Let T (G, d) be the set of trees
in G which have depth d and for which every internal vertex has r children. For a tree
T ∈ T (G, d) let AT be the event that all leaves in T are initially infected. Note that this is an
event on the product space of initial infection, where every vertex is infected independently
with probability p. Also, note that this is a non-decreasing event: if we infect more vertices,
then AT will not stop holding.
T
c
We wish to show P
A
T ∈T (G,d) T > 0. To this end, we apply the FKG inequality (see
for example Theorem 6.3.2 in [27]):
\
Y
P(
AcT ) ≥
(1 − P(AT ))
T ∈T (G,d)
T ∈T (G,d)

≥ exp −2

X
P(AT ) ,
(39)
T ∈T (G,d)
where the last inequality follows as 1 − x ≥ e−2x when x is small enough. In this case, it will
be small enough provided that t is large, since P(AT ) = p` where ` is the number of leaves in
T and p = o(1).
33
Let T (G, d, `) ⊆ T (G, d) be those depth-d trees in G with ` leaves. We have
X
X X
X
p` =
p` |T (G, d, `)|
P(AT ) =
T ∈T (G,d)
`≥d T ∈T (G,d,`)
`≥d
Let C be some large constant, let σ(d, `) = {G ∈ PAt (m, δ) : |T (d, `)| ≤ CE[|T (d, `)|]}
where |T (d, `)| is the random variable on PAt (m, δ) that counts the number
of depth-d trees
T
with ` leaves and each internal vertex having r children. Let σ(d) = `≥d σ(d, `). Then if
G ∈ σ(d)
X
X
P(AT ) ≤ C
p` E[|T (d, `)|] = O(1),
T ∈T (G,d)
`≥d
where the last equality follows from (22) when d = d0 , replacing ω in p = pc /ω (which gave
us o(1)) with 1/λ.
Now
1
P(|T (PAt (m, δ), d, `)| > CE[|T (d, `)|]) ≤ .
C
d+1
d+1
Hence P(PAt (m, δ) ∈
/ σ(d)) ≤ r /C since d ≤ ` ≤ r . Of course, we choose C > rd+1 .
Getting back to (39), we see that with probability at least 1 − rd+1 /C,


\
X
P(
AcT |G ∈ σ(d)) ≥ exp −2
P(AT ) = Ω(1).
T ∈T (G,d)
T ∈T (G,d)
Consequently, with probability at least p1 > 0, there is no witness tree of depth d = d0 ,
meaning no infection occurs in this round or thereafter.
The same argument applies to witness structures which are not trees. As per above, their
expected number of occurrences is bounded from above by that of witness trees.
When d < d0 , the results of the previous section show that the expected number of
infected vertices in round d > 0 is o(t1−γ ). Hence, the above analysis together with Markov’s
inequality yields |If |/|I0 | < 1 + ε, for ε > 0, with probability at least p1 > 0, for any t large
enough.
Proof of Theorem 3(ii). We wish to show there is a full outbreak. This will happen if, for
some k ≥ 1, the first k vertices [k] get infected, and additionally, no vertex has more than one
self-loop. We will show that this happens with some probability bounded away from zero.
Consider vertex 1. The argument is along the following lines: The expected degree of
vertex 1 is about tγ . Suppose that the actual degree of vertex 1 is roughly its expected
degree. When the infection probability is p = λ/tγ where λ is a constant, then the probability
of vertex 1 getting infected in round τ = 1 is P(Bin(tγ , λ/tγ ) ≥ r), which is asymptotically
bounded away from 0.
For δ ≥ 0, we can use Lemma 5. Setting h > 0 to be a sufficiently small constant, we get
P(S1 (t) < E[S1 (t)]/K) < 1/eh < 1. Since S1 (t) = D1 (t) and so P(D1 (t) ≥ tγ ) ≥ 1 for some
constants , 1 > 0. For δ < 0 we apply Lemma 6 with i = 1 to get the same
T result.
Let Ei be the event that vertex i has at most one self-loop and let E = i>1 Ei . Let A be
the event D1 (t) ≥ tγ . It is clear that P(A ∩ E) ≥ P(A )P(E).
As per the previous sections, for i > 1, P(Eic ) = O(1/i2 ) and so lim inf t→∞ P(E) > 0.
Therefore, with some probability bounded away from zero, no vertex has more than one self
loop, and vertex 1 is infected in round τ = 1. Consequently, all vertices become infected
eventually.
34
7
Conclusions - open questions
This paper studies the evolution of a bootstrap percolation process on random graphs that
have been generated through preferential attachment and generalise the classical BarabásiAlbert model. For r < m, where 2m is the average degree, we determine a critical function
ac (t) such that when the size a(t) of the initial set “crosses” ac (t) the evolution of the bootstrap
percolation process with activation threshold r changes abruptly from almost no evolution to
full infection. The critical function satisfies ac (t) = o(t), which implies that a sublinear initial
infection leads to full infection.
Our results are somewhat less tight for r = 2. It would be interesting to find out whether
the sharpness of the threshold that we deduced for r ≥ 3 also holds in this case. Also,
the critical window itself for the case r = 2 has not been explored in the present work.
Furthermore, it would be interesting to determine the number of rounds until the complete
infection of all vertices in the supercritical case.
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8
Appendix
8.1
Proofs
Proof of Lemma 5. Assume h, ct , A > 0. Recall also that here we assume that δ ≥ 0. Let
Zt = Si (t).
−hZ
−hA
t
P (Zt < A) = P e ct > e ct
.
Zt−1
Zt = Zt−1 + Yt . Then Yt Xt ∼ Bin m, mt(2+δ/m)
.
E e
where p =
−hXt
ct
−h m
| Zt−1 = 1 − p + pe ct
Zt−1
mt(2+δ/m) .
37
Using e−x ≤ 1 − x + x2 ,
1 − p + pe
−h
ct
m
≤
=
≤
=
2 !m
h
h
1−p+p−p +p
ct
ct
m
h
h
1−
1−p
ct
ct
mph
h
exp −
1−
ct
ct
h
hZt−1
1−
exp −
ct (2 + δ/m) t
ct
Then
E e
−hZt−1
ct
e
−hYt
ct
| Zt−1
≤ exp −
hZt−1
ct (2 + δ/m) t
h
hZt−1
1−
−
.
ct
ct
Taking expectations on both sides,
hZt−1
1 − h/ct
−hZt
≤ E exp −
1+
.
E exp
ct
ct
(2 + δ/m)t
1
Let ci = 1 and ct = 1 + γt ct−1 = 1 + (2+δ/m)t
ct−1 for t > i, and note that ct ∼
We have,



1−h/ct
1
+
−hZt
hZt−1
(2+δ/m)t 
E exp
≤ E exp −
1
ct
ct−1 1 + (2+δ/m)t
hZt−1
h
≤ E exp −
.
1−
ct−1
(2 + δ/m)ct t
Iterating,
E exp
−hZt
ct
hZt−1
hγ
≤ E exp −
1−
ct−1
ct t
hZt−2
hγ
hγ
≤ E exp −
1−
1−
ct−2
ct t
ct−1 (t − 1)
..
.



t Y
hZ
hγ
i

≤ E exp −
1−
ci
cj j
j=i



t Y
hγ 
= E exp −2hmi
1−
.
cj j
j=i
38
t γ
i .
t t
Y
X
hγ
1
1−
≥ 1 − hγ
cj j
jcj
j=i
j=i


t
X
1
γ 
= 1 − O h
j
j=i j i
= 1 − O hiγ i−γ
= 1 − O (h) .
So
E exp
−hZt
ct
≤ E [exp (−2hmi (1 − O (h)))] = exp (−2hmi (1 − O (h))) .
Hence using Markov’s inequality,
−hZ
−hA
t
e−2hmi(1−O(h))
c
c
P e t >e t
.
≤
−hA
e ct
γ
Recalling that ict ∼ i ti = tγ i1−γ and E[Si (t)] ≥ C` tγ i1−γ (cf. Proposition 4), choose
√
a sufficiently
large
constant
constant
K
such
that
E[S
(t)]/K
<
C
ic
/
K and let A =
i
t
`
√
C` ict / K. Then,
−hZ
−hA
t
1
c
c
≤ P e t >e t
P Si (t) ≤ E[Si (t)]
K
√ ≤ exp −2hmi (1 − O (h)) + hiC` / K
√ = exp −hi 2m − O(h) − C` / K
≤ exp (−hi) ,
where the last inequality follows if K > K0 where K0 > 0 is a sufficiently large constant that
need only depend on m, δ, and if h is small enough.
Proof of Lemma 15. The first sum is bounded from above by an integral:
t
X
j 0 =j+1
1
1
fa0 (j ) γ 01−γ ≤ γ
j j
j
0
Z
t
va0 (x)
j
1
x1−γ
dx.
Assume that xa0 is a contraction leaf. In this case, the above integral becomes
Z t
Z t
1
(log x)ρa0
va0 (x) 1−γ dx =
dx.
1−γ+ea0
x
j
j x
The value of this integral now depends on the sign of −γ + ea0 . Recall that ea0 > 0 and it
satisfies Property (A). Assume that ea0 = Aγ + B(1 − γ). Since 1 − γ ≥ γ, the latter is
ea0 ≥ (A + B)γ ≥ rγ. Since r > 1, it follows that −γ + ea0 > (r − 1)γ > 0. Hence, by
Lemma 17
Z t
1
va0 (x) 1−γ dx . (log j)ρa0 j γ−ea0 ,
x
j
39
and therefore
t
X
fa0 (j 0 )
j 0 =j+1
1
. (log j)ρa0 j −ea0 .
j γ j 01−γ
Assume now that xa0 is an original leaf. In this case,
t
X
j 0 =j+1
1
1
fa0 (j ) γ 01−γ . γ
j j
j
0
Z
j
t
1
1
va0 (x) 1−γ dx = p γ
x
j
Z
j
t
1
x1−γ
dx . p
1 γ
1 1
t =
.
γ
j
ω jγ
Consider now the second sum in the statement of the Lemma. If xa0 is an original leaf,
then
j−1
X
j 0 =1
fa0 (j 0 )
1
1
. 1−γ
0γ
1−γ
j j
j
j
Z
va0 (x)
1
1
1
dx = p 1−γ
γ
x
j
Z
j
1
j≤t 1 1
1
1
1
dx . γ 1−γ j 1−γ ≤
.
γ
x
ωt j
ω jγ
If xa0 is a contraction leaf, then we have
j−1
X
j 0 =1
1
1
fa0 (j ) 0γ 1−γ . 1−γ
j j
j
0
Z
j
1
1
1
va0 (x) γ dx = 1−γ
x
j
Z
1
j
(log x)ρa0
dx.
xγ+ea0
If 1 − γ > ea0 , then the above becomes
j−1
X
j 0 =1
fa0 (j 0 )
1
.
j 0γ j 1−γ
1
j 1−γ
(log j)ρa0 j 1−γ−ea0 = (log j)ρa0 j −ea0 .
If 1 − γ ≤ ea0 , then we will get
j−1
X
fa0 (j 0 )
j 0 =1
8.2
1
(log j)ρa0 +1
.
.
j 0γ j 1−γ
j 1−γ
Proof of Lemma 17
In this section, we prove the following lemma 17, which has been useful for our calculations.
We restate it here for completeness.
Lemma 21. Let k ≥ 0 be an integer, let α > 0 be a real number and let
Z t
Ik,α (j) :=
(log x)k x−1−α dx.
j
Then uniformly for j ≥ 1 we have
Ik,α (j) .
(log j ∨ 1)k
.
jα
40
Proof. Let v = (log x)k , meaning
Integration by parts gives
Ik,α (j) =
kx
−(log x)
dv
dx
=
−α t
α
j
k(log x)k−1
.
x
k
+
α
Z
t
Let
du
dx
(log x)k−1 x−1−α dx
j
1)k
1 (log j ∨
k
+ Ik−1,α
α
jα
α
k 1 (log j ∨ 1)k−1 k − 1
1 (log j ∨ 1)k
+
+
≤
Ik−2,α
α
jα
α α
jα
α
k (log j ∨ 1)k−1 k(k − 1)
1 (log j ∨ 1)k
+
+
Ik−2,α
=
α
jα
α2
jα
α2
k (log j ∨ 1)k−1 k(k − 1) (log j ∨ 1)k−2
1 (log j ∨ 1)k
+
+
+
≤
α
jα
α2
jα
α3
jα
k(k − 1) . . . 2 log j ∨ 1 k(k − 1) . . . 2
... +
+
I0,α (j).
jα
αk
αk
≤
Now
Z
I0,α (j) =
j
t
x−1−α dx ≤
1
.
αj α
Thus, we get
Ik,α (j) ≤
.
1 (log j ∨ 1)k
k (log j ∨ 1)k−1
k(k − 1) . . . 2 log j ∨ 1
k! 1
+
+ ... +
+ k+1 α
α
2
α
α
k
α
j
α
j
j
j
α
α
k
(log j ∨ 1)
,
jα
uniformly over all j ≥ 1.
41
−α
= x−1−α , meaning u = − x α .